Ｃａｌｃｕｌｕｓ　１　ｌｅｃｔｕｒｅ　１．２

Pｒｏｐｅｒｔｉｅｓ　Oｆ　Lｉｍｉｔｓ．　Ｔｅｃｈｎｉｑｕｅｓ　Oｆ　Lｉｍｉｔ　Cｏｍｐｕｔａｔｉｏｎ

Ｉｎｔｒｏｄｕｃｔｉｏｎ ｔｏ ｔｈｅ Ｐｒｏｐｅｒｔｉｅｓ ｏｆ Ｌｉｍｉｔｓ

• 0:50. The Limit of a Constant: – [📷image]
  • The limit of a constant is defined as: $\displaystyle \lim_{x\to a} 𝓒=𝓒$, where $𝓒$ is a constant.
  • If a function always has the same value, it doesn’t matter what value $x$ approaches — the result does not change.
• 2:53. The Limit of a Simple Variable: – [📷image]
  • The limit of a simple variable is defined as: $\displaystyle \lim_{x\to a} 𝒙=𝒂$.
  • If the function is just $𝒙$ itself, then as $𝒙$ gets closer to $a$, the function gets closer to $a$ — so the limit is the value it approaches.
• 5:10. The Importance of Equality of One-sided Limits: – [📷image]
  • The importance that the one-sided limits are equal for the general limit to e𝒙ist is emphasized.
  • Examples of one-sided limits that are not equal are provided, resulting in the non-e𝒙istence of the general limit.
    • $\displaystyle \lim_{x\to 0^-}\dfrac{1}{x}=-\infty$ and $\displaystyle \lim_{x\to 0^+}\dfrac{1}{𝒙}=+\infty$

Ｐｒｏｐｅｒｔｉｅｓ ｏｆ Ｌｉｍｉｔｓ

• 6:25. Properties of Limits: – [📷image]
  • The properties of limits are introduced, including sum, difference, product, quotient, and e𝒙ponents.
    • If $\displaystyle \lim_{x\to a}𝒇(𝒙)=L$ and $\displaystyle \lim_{x\to a}𝓰(𝒙)=L_{2}$, then:
      • 1.2.1 $\displaystyle \lim_{x\to a}\big(𝒇(𝒙)\pm 𝓰(𝒙)\big)=\lim_{x\to a} 𝒇(𝒙)\ \pm\ \lim_{x\to a} 𝓰(𝒙)$
      • 1.2.2 $\displaystyle \lim_{x\to a}\big(𝒇(𝒙)\cdot 𝓰(𝒙)\big)=\big(\lim_{x\to a} 𝒇(𝒙)\big)\cdot \big(\lim_{x\to a} 𝓰(𝒙)\big)$
      • 1.2.3 $\displaystyle \lim_{x\to a}\dfrac{𝒇(𝒙)}{𝓰(𝒙)}=\dfrac{\lim_{x\to a} 𝒇(𝒙)}{\lim_{x\to a} 𝓰(𝒙)}$, provided $\displaystyle \lim_{x\to a} 𝓰(𝒙)\neq 0$
      • 1.2.4 $\displaystyle \lim_{x\to a}\big(𝒇(𝒙)\big)^{n}=\big(\lim_{x\to a} 𝒇(𝒙)\big)^{n}$
        • ▣ 1.2.4.1 $\displaystyle \lim_{x\to a}\sqrt[n]{\,𝒇(𝒙)\,}=\sqrt[n]{\,\lim_{x\to a} 𝒇(𝒙)\,}$ (✔ The property is valid only if the limit exists, and in the case of even roots, the value inside the root must be non-negative)
• 11:05. Recapitulation

Ｅｖａｌｕａｔｉｎｇ Ｌｉｍｉｔｓ ｉｎ Ｐｏｌｙｎｏｍｉａｌｓ

• 12:20. 🧩 Evaluation of $\displaystyle \lim_{x\to 2}\ (𝒙^{3}-2𝒙+7)$: – [📷image]
  • $\displaystyle \lim_{x\to 2}(𝒙^{3}-2𝒙+7)$ is calculated using the properties of limits.
    • $\displaystyle \lim_{x\to 2}𝒙^{3}-\lim_{x\to 2}2𝒙+\lim_{x\to 2}7$
    • $\displaystyle \lim_{x\to 2}𝒙^{3}-\big(\lim_{x\to 2}2\big)\cdot \big(\lim_{x\to 2}𝒙\big)+\lim_{x\to 2}7$
    • $\displaystyle \lim_{x\to 2}𝒙^{3}-2𝒙+7=11$
    • $f(2)=11$
• 17:04. 𝑻𝒉𝒆 𝑳𝒊𝒎𝒊𝒕 𝒐𝒇 𝑨𝒏𝒚 𝑷𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍: – [📷image]
  • $\displaystyle \lim_{x\to a}\mathcal{P}(𝒙)=\mathcal{P}(𝒂)$
    • 𝑻𝒉𝒆 𝒍𝒊𝒎𝒊𝒕 𝒐𝒇 𝒂 𝒑𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒂𝒔 $𝒙$ 𝒂𝒑𝒑𝒓𝒐𝒂𝒄𝒉𝒆𝒔 𝒂 𝒑𝒐𝒊𝒏𝒕 $a$ 𝒊𝒔 𝒋𝒖𝒔𝒕 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍 𝒂𝒕 𝒕𝒉𝒂𝒕 𝒑𝒐𝒊𝒏𝒕.
• 20:16. 🧩 Evaluation of $\displaystyle \lim_{x\to 2}\ \big(𝒙^{5}-3𝒙+4\big)^{3}$: – [📷image]
  • The calculation is simplified by evaluating the function at $𝒙=2$ and then cubing the result.

Ｌｉｍｉｔｓ ｏｆ Ｎｏｎ－Ｐｏｌｙｎｏｍｉａｌ Ｆｕｎｃｔｉｏｎｓ

• 23:05. 🧩 Evaluation of $\displaystyle \lim_{x\to 2}\ \dfrac{4𝒙^{2}+1}{𝒙+3}$: – [📷image]
  • Apply property 1.2.3 → $\displaystyle \dfrac{\lim_{x\to 2}(4𝒙^{2}+1)}{\lim_{x\to 2}(𝒙+3)}$
  • Then we check that both the numerator and the denominator are polynomials, so we can substitute $𝒙=2$ directly.
• 26:20. 🧩 Evaluation of $\displaystyle \lim_{x\to 1}\ \sqrt[3]{\dfrac{5𝒙+7}{𝒙^{2}+1}}$ – [📷image]
  • $\displaystyle \lim_{x\to 1}\sqrt[3]{\dfrac{5𝒙+7}{𝒙^{2}+1}}=\sqrt[3]{\dfrac{5\cdot 1+7}{1^{2}+1}}=\sqrt[3]{6}$
  • It is mentioned that the same properties as for polynomials can be used, as long as domain issues, such as division by zero, are avoided.
• 28:50. 🧩 Evaluation of $\displaystyle \lim_{x\to 2}\ \dfrac{𝒙^{2}-4}{𝒙-2}$ – [📷image]
  • $\displaystyle \lim_{x\to 2}\dfrac{𝒙^{2}-4}{𝒙-2}$
  • $\displaystyle \lim_{x\to 2}\dfrac{(𝒙+2)(𝒙-2)}{𝒙-2}$
  • $\displaystyle \lim_{x\to 2}\big[\,𝒙+2\,\big]$
  • Cancel $(𝒙-2)$, since we are not evaluating exactly at $𝒙=2$, but rather approaching it.
  • $\displaystyle \lim_{x\to 2}\big[𝒙+2\big]=2+2=4$
  • The original function is undefined because the denominator becomes zero $(𝒙-2=0)$.
    • This creates a 𝒉𝒐𝒍𝒆 in the graph at the point $(2,4)$. The function does not diverge to infinity; it simply has a removable discontinuity.
    • This is different from a vertical asymptote, which occurs when the denominator approaches zero but the numerator does 𝒏𝒐𝒕 approach zero at the same time (the numerator doesn’t become zero when the denominator does) — causing the function to diverge to $+\infty$ or $-\infty$.
• 📝 N͟O͟T͟E͟: 𝑰𝒇 𝒕𝒉𝒆 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 𝒛𝒆𝒓𝒐 𝒂𝒏𝒅 𝒘𝒆 𝒄𝒂𝒏 𝒔𝒊𝒎𝒑𝒍𝒊𝒇𝒚 𝒕𝒉𝒆 𝒆𝒙𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 𝒃𝒚 𝒄𝒂𝒏𝒄𝒆𝒍𝒊𝒏𝒈 𝒂 𝒄𝒐𝒎𝒎𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓 𝒘𝒊𝒕𝒉 𝒕𝒉𝒆 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓, 𝒕𝒉𝒆𝒏 𝒕𝒉𝒆𝒓𝒆'𝒔 𝒋𝒖𝒔𝒕 𝒂 𝒉𝒐𝒍𝒆 — 𝒕𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒊𝒔 𝒔𝒕𝒊𝒍𝒍 𝒘𝒆𝒍𝒍-𝒃𝒆𝒉𝒂𝒗𝒆𝒅 𝒏𝒆𝒂𝒓𝒃𝒚. 𝑩𝒖𝒕 𝒊𝒇 𝒕𝒉𝒆 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 𝒊𝒔 𝒛𝒆𝒓𝒐 𝒂𝒏𝒅 𝒘𝒆 𝒄͟𝒂͟𝒏͟’͟𝒕͟ 𝒔𝒊𝒎𝒑𝒍𝒊𝒇𝒚, 𝒕𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒃𝒍𝒐𝒘𝒔 𝒖𝒑 — 𝒕𝒉𝒂𝒕'𝒔 𝒂 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒂𝒔𝒚𝒎𝒑𝒕𝒐𝒕𝒆.

Ｆａｃｔｏｒｉｚａｔｉｏｎ ａｎｄ Ｓｉｍｐｌｉｆｉｃａｔｉｏｎ ｔｏ Ｅｖａｌｕａｔｅ Ｌｉｍｉｔｓ

• 33:45. Factorization and Simplification to Evaluate Limits:
  • 🧩 Evaluation of $\displaystyle \lim_{x\to -4}\ \dfrac{2𝒙+8}{𝒙^{2}+𝒙-12}$ – [📷image]
    • $\displaystyle \lim_{x\to -4}\dfrac{2𝒙+8}{𝒙^{2}+𝒙-12}$
    • $\displaystyle \lim_{x\to -4}\dfrac{2(𝒙+4)}{(𝒙+4)(𝒙-3)}$
    • $\displaystyle \lim_{x\to -4}\dfrac{2}{𝒙-3}= \dfrac{2}{-4-3}=\dfrac{2}{-7}=-\dfrac{2}{7}$
  • The technique of factoring and simplifying serve for evaluate limits when direct substitution results in $0/0$.
  • 𝑪𝒐𝒎𝒎𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓𝒔 𝒊𝒏 𝒂 $0/0$ 𝒊𝒏𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒏 𝒇𝒐𝒓 𝒑𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍𝒔: 𝑰𝒇 𝒃𝒐𝒕𝒉 𝒕𝒉𝒆 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓 𝒂𝒏𝒅 𝒕𝒉𝒆 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 𝒂𝒓𝒆 𝒑𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍𝒔 𝒂𝒏𝒅 𝒃𝒆𝒄𝒐𝒎𝒆 $0$ 𝒂𝒕 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝒑𝒐𝒊𝒏𝒕 $x=c$, 𝒊𝒕 𝒎𝒆𝒂𝒏𝒔 $x=c$ 𝒊𝒔 𝒂 𝒓𝒐𝒐𝒕 𝒐𝒇 𝒃𝒐𝒕𝒉 𝒑𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍𝒔. 𝑻𝒉𝒊𝒔 𝒊𝒎𝒑𝒍𝒊𝒆𝒔 𝒕𝒉𝒆𝒚 𝒔𝒉𝒂𝒓𝒆 𝒂 𝒄𝒐𝒎𝒎𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓 $(𝒙-𝒄)$. 𝑻𝒉𝒊𝒔 𝒇𝒂𝒄𝒕𝒐𝒓 𝒄𝒂𝒏 𝒃𝒆 𝒆𝒍𝒊𝒎𝒊𝒏𝒂𝒕𝒆𝒅 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒔𝒊𝒎𝒑𝒍𝒊𝒇𝒊𝒄𝒂𝒕𝒊𝒐𝒏, 𝒂𝒍𝒍𝒐𝒘𝒊𝒏𝒈 𝒕𝒉𝒆 𝒍𝒊𝒎𝒊𝒕 𝒕𝒐 𝒃𝒆 𝒓𝒆𝒔𝒐𝒍𝒗𝒆𝒅.
  • The importance of maintaining the limit notation until the limit is evaluated is emphasized.

Ｌｉｍｉｔｓ ｗｉｔｈ Ｖｅｒｔｉｃａｌ Ａｓｙｍｐｔｏｔｅｓ

• 38:23. 🧩 Evaluation of $\displaystyle \lim_{x\to 5}\ \dfrac{𝒙^{2}-3𝒙-10}{𝒙^{2}-10𝒙+25}$ – [📷image]
  • $\displaystyle \lim_{x\to 5}\dfrac{𝒙^{2}-3𝒙-10}{𝒙^{2}-10𝒙+25}$ is attempted by factoring and simplifying, but a 🔶vertical asymptote🔶 is found.
  • $\displaystyle \lim_{x\to 5}\dfrac{(𝒙-5)(𝒙+2)}{(𝒙-5)(𝒙-5)}$ = $\displaystyle \lim_{x\to 5}\dfrac{𝒙+2}{𝒙-5}$
  • 📝 N͟O͟T͟E͟: Before rushing to factor the expression, 𝒂𝒍𝒘𝒂𝒚𝒔 𝒄𝒉𝒆𝒄𝒌 the value you're approaching—in this case, $𝒙=5$. If plugging in the number gives a defined result, there's no need to factor. Factoring would just waste time and might confuse you unnecessarily.
  • The difference is clarified between:
    • A 𝒉𝒐𝒍𝒆 in the graph (removable discontinuity, can be simplified), and
    • A 🔶vertical asymptote🔶 (non-removable, cannot be simplified).
  • 40:45. ͟𝐒͟𝐢͟𝐠͟𝐧͟ ͟𝐀͟𝐧͟𝐚͟𝐥͟𝐲͟𝐬͟𝐢͟𝐬͟ ͟𝐓͟𝐞͟𝐬͟𝐭͟:
    • The sign analysis test is presented to determine the behavior of a function around a 🔶vertical asymptote🔶.
    • Steps for Sign Analysis at $𝒙=5$:
      • 1. Check the function:
        • At $𝒙=5$, the denominator becomes $0$ and cannot be canceled out.
        • This means there is a 🔶vertical asymptote🔶 at $𝒙=5$ — not a hole.
      • 2. Identify key points:
        • Problem point: $𝒙=5$
        • Optional helper point: $𝒙=-2$ (just to clearly separate intervals on the number line)
      • 3. Draw a number line to analyze signs
        
      • 4. Pick a number to the right of 5 (e.g., $𝒙=6$):
        • $𝒇(𝒙)=\dfrac{𝒙+2}{𝒙-5}$
        • $𝒇(6)=\dfrac{6+2}{6-5}=\dfrac{8}{1}=8>0$ → Positive
        • So as $𝒙\to 5^{+}$, $𝒇(𝒙)\to +\infty$
      • 5. Pick a number to the left of 5 (e.g., $𝒙=0$):
        • $𝒇(𝒙)=\dfrac{𝒙+2}{𝒙-5}$
        • $𝒇(0)=\dfrac{0+2}{0-5}=\dfrac{2}{-5}=-\dfrac{2}{5}<0$ → Negative
        • So as $𝒙\to 5^{-}$, $𝒇(𝒙)\to -\infty$
      • 6. The limit D.N.E because the left-hand limit ($-\infty$) and right-hand limit ($+\infty$) are not the same. This confirms a vertical asymptote at $𝒙=5$.

Ｓｕｍｍａｒｙ ｏｆ Ｔｅｃｈｎｉｑｕｅｓ ｆｏｒ Ｅｖａｌｕａｔｉｎｇ Ｌｉｍｉｔｓ

• 47:35. Summary of Techniques for Evaluating Limits
  • 📝 N͟O͟T͟E͟:
    • ① If $0/0$, factor and simplify.
    • ② If you can’t cancel the “problem,” check with ͟𝐒͟𝐢͟𝐠͟𝐧͟ ͟𝐀͟𝐧͟𝐚͟𝐥͟𝐲͟𝐬͟𝐢͟𝐬͟ ͟𝐓͟𝐞͟𝐬͟𝐭͟ because the limit might not exist.

Ｒａｔｉｏｎａｌｉｚａｔｉｏｎ ｔｏ Ｅｖａｌｕａｔｅ Ｌｉｍｉｔｓ

• Rationalization to Evaluate Limits:
  • When solving limits involving radicals:
    • ① Multiply by the conjugate:
      • Especially if there’s a radical in the denominator.
      • The goal is to eliminate the radical from the denominator. Multiply by the conjugate to use the identity:
      • $(a-b)(a+b)=a^{2}-b^{2}$
      • This clears the radical in the denominator.
    • ② Do not expand (distribute) the numerator immediately:
      • Avoid expanding the numerator right away.
      • First, look to simplify — sometimes you can cancel factors.
      • If you expand too soon, you might miss cancellations and make the problem messier.
• 50:50. 🧩 Evaluation of $\displaystyle \lim_{x\to 1}\ \dfrac{𝒙-1}{\sqrt{𝒙}-1}$ – [📷image]
  • $\displaystyle \lim_{x\to 1}\dfrac{𝒙-1}{\sqrt{𝒙}-1}$
  • $\displaystyle \lim_{x\to 1}\dfrac{(𝒙-1)\cdot(\sqrt{𝒙}+1)}{(𝒙-1)}$ Multiply by the conjugate: $\dfrac{\sqrt{𝒙}+1}{\sqrt{𝒙}+1}$
  • $\displaystyle \lim_{x\to 1}\big(\sqrt{𝒙}+1\big)$ Simplify cancelling $(𝒙-1)$
  • $\sqrt{1}+1=2$ Substitute directly
• 55:25. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{\sqrt{1+𝒙}-1}{𝒙}$ – [📷image]
  • $\displaystyle \lim_{x\to 0}\dfrac{\sqrt{1+𝒙}-1}{𝒙}$
  • $\displaystyle \lim_{x\to 0}\dfrac{(1+𝒙)-1}{\,𝒙\cdot(\sqrt{1+𝒙}+1)\,}$ Multiply by the conjugate: $\dfrac{\sqrt{1+𝒙}+1}{\sqrt{1+𝒙}+1}$
  • $\displaystyle \lim_{x\to 0}\dfrac{1}{\sqrt{1+𝒙}+1}$ Simplify cancelling $𝒙$
  • $\dfrac{1}{\sqrt{1+0}+1}=\dfrac{1}{2}$ Substitute directly

Ｌｉｍｉｔｓ ｂｙ Ｐａｒｔｓ

• 1:01:10. Limits by Parts
  • The concept of limits by parts is introduced, where different functions are considered for different intervals of the domain.
  • I͟D͟E͟A͟: To evaluate a limit by parts (limits for piecewise functions), the one-sided limits 🟪at the points '𝒄' of change of the functions must be considered🟪 and see if they're equal:
    • ① You must check the limit from the left $(𝒙\to 𝒄^{-})$ and from the right $(𝒙\to 𝒄^{+})$ at the point $𝒄$.
      • Finding the limit from the left means identifying which piece of the function approaches $𝒄$ from values smaller than $𝒄$.
      • Finding the limit from the right means identifying which piece of the function approaches $𝒄$ from values larger than $𝒄$.
    • ② If both one-sided limits are equal, then the limit exists at $𝒄$. If they are different, the overall limit does not exist at that point.
• 1:03:12. 🧩 Evaluation of Limits by Parts: – [📷image-1] – [📷image-2]
  • The need to determine which function to use for each one-sided limit, depending on the interval of the domain, is emphasized.
  • An e𝒙ample of limit by parts is worked out for a function defined in three intervals. The function is:
    • $\displaystyle𝒇(𝒙)=\begin{cases}\dfrac{1}{𝒙+2}, & \text{if } 𝒙<-2 \quad (𝒇_{1})\\[6pt]𝒙^{2}-5, & \text{if } -2<𝒙\le 3 \quad (𝒇_{2})\\[6pt]\sqrt{𝒙+13}, & \text{if } 𝒙>3 \quad (𝒇_{3})\end{cases}$
  • 🟪Points '𝒄' of Change: find the limits as $𝒙$ approaches $-2$ and $3$🟪
  • 1:05:10. Graphical Representation the e𝒙ample.
  • In this piecewise function $𝒇(𝒙)$:
    • 🟪We must check the limit at each transition point $(𝒙=-2 \text{ and } 𝒙=3)$🟪. Steps:
      • 1. Find the limit from the left $(𝒙\to 𝒄^{-})$ and the right $(𝒙\to 𝒄^{+})$ separately at the point $𝒄$:
        • Finding the limit from the left means identifying which piece of the function is used when approaching $𝒄$ from the left.
        • Example: For $𝒙\to -2^{-}$, use $𝒇_{1}(𝒙)$.
        • Example: For $𝒙\to 3^{-}$, use $𝒇_{2}(𝒙)$.
        • Finding the limit from the right means identifying which piece of the function is used when approaching $𝒄$ from the right.
        • Example: For $𝒙\to -2^{+}$, use $𝒇_{2}(𝒙)$.
        • Example: For $𝒙\to 3^{+}$, use $𝒇_{3}(𝒙)$.
      • 2. Compare the limits at each point:
        • $𝒙=3$
        • $\displaystyle \lim_{x\to 3^{-}}\big(𝒙^{2}-5\big)=4$
        • $\displaystyle \lim_{x\to 3^{+}}\sqrt{𝒙+13}=4$
        • $\displaystyle \lim_{x\to 3}𝒇(𝒙)=4$
        • Due to the left and right limits are equal, the limit exists at $𝒙=3$.
        • $𝒙=-2$
        • $\displaystyle \lim_{x\to -2^{-}} \dfrac{1}{𝒙+2}=-\infty$
        • $\displaystyle \lim_{x\to -2^{+}} 𝒙^{2}-5=-1$
        • $\displaystyle \lim_{x\to -2}𝒇(𝒙)=\text{D.N.E.}$ (Does Not Exist)
        • Due to they are different, the limit does not exist at $𝒙=-2$.

Ｌｉｍｉｔｓ ｉｎ Ｔｒｉｇｏｎｏｍｅｔｒｉｃ Ｆｕｎｃｔｉｏｎｓ

• 1:19:43. 𝙨𝒊𝙣(𝒙) 𝒂𝙣𝒅 𝒄𝙤𝒔(𝒙) 𝙖𝙧𝙚 𝙘𝙤𝙣𝙩𝙞𝙣𝙪𝙤𝙪𝙨 𝒆𝙫𝒆𝙧𝙮𝙬𝒉𝙚𝙧𝙚. – [📷image]
  • $\displaystyle \lim_{x\to a}\sin(𝒙)=\sin(𝒂)$
  • $\displaystyle \lim_{x\to a}\cos(𝒙)=\cos(𝒂)$
• 1:22:32. 𝑻𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 $\tan(𝒙)$ 𝒆𝒙𝒊𝒔𝒕𝒔 𝒆𝒙𝒄𝒆𝒑𝒕 𝒂𝒕 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕𝒔 𝒘𝒉𝒆𝒓𝒆 $\cos(𝒙)=0$. 𝑻𝒉𝒆𝒔𝒆 𝒑𝒐𝒊𝒏𝒕𝒔 𝒂𝒓𝒆 $𝒙=\pm \pi/2+n\pi,\ n\in\mathbb{Z}$. – [📷image]
  • $\displaystyle \lim_{x\to a}\tan(𝒙)=\tan(a)$ ; $𝒙\neq \pm \pi/2+n\pi,\ n\in\mathbb{Z}$.
• 1:28:15. 🧩 Evaluation of $\displaystyle \lim_{x\to 1}\ \cos\!\left(\dfrac{𝒙^{2}-1}{𝒙-1}\right)$ – [📷image]
  • $\cos(𝒙)$ is continuous so by composition ...
  • $\displaystyle \cos\!\left(\lim_{x\to 1}\dfrac{𝒙^{2}-1}{𝒙-1}\right)$
  • $\displaystyle \cos\!\left(\lim_{x\to 1}\dfrac{(𝒙+1)(𝒙-1)}{𝒙-1}\right)$
  • $\displaystyle \cos\!\big(\lim_{x\to 1}(𝒙+1)\big)$
  • $\cos(1+1)$
  • $\cos(2)$
  • 📝N͟O͟T͟E͟: Since $\cos(𝒙)$ is continuous, we can move the limit inside by the composition rule:
    • 𝓟𝓻𝓸𝓹𝓮𝓻𝓽𝔂: Limit of Continuous Functions (Composition Rule)
      • Statement:
        • If $𝒇(𝒙)$ is continuous at $𝒙=a$, then:
          • $\displaystyle \lim_{x\to a}𝒇(𝒙)=𝒇(a)$
      • Application:
        • For continuous functions like $\sin(𝒙)$, $\cos(𝒙)$, polynomials, rationals, exponentials, etc., you can move the limit inside:
          • $\displaystyle \lim_{x\to a}𝒇(𝒙)=𝒇\!\big(\lim_{x\to a} 𝒙\big)=𝒇(a)$
      • Important:
        • This is why, when dealing with continuous functions, limits are straightforward to compute.
  • 1:33:00. 🧩 Evaluation of $\displaystyle \lim_{x\to \pi/2}\ \big(3𝒙^{2}+\cos(𝒙)\big)$ – [📷image]
    • Since there is no issue of discontinuity or undefined values at $𝒙=\pi/2$, we are allowed to directly substitute the value into the expression.
    • 📝N͟O͟T͟E͟: Whenever the function is continuous at the point — meaning there is no division by zero, no square root of a negative number in the real domain, and no discontinuity — we can evaluate the limit by direct substitution.
    • So: $\displaystyle \lim_{x\to \pi/2}\ \big(3𝒙^{2}+\cos(𝒙)\big)=3(\pi/2)^{2}+\cos(\pi/2)=3\cdot \dfrac{\pi^{2}}{4}+0=\dfrac{3\pi^{2}}{4}$
  • 1:37:15. ∴ 𝑫𝒆𝒎𝒐𝒏𝒔𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 $\displaystyle \lim_{x\to 0}\ \dfrac{\sin 𝒙}{𝒙}=1$ – [📷image-1] – [📷image-2]
    • $\displaystyle \lim_{x\to 0}\dfrac{\sin 𝒙}{𝒙}=1$ is demonstrated using the squeeze theorem and areas of triangles and sectors.
    • Triangles and sectors in a unit circle are constructed to compare their areas and establish the necessary inequalities for the squeeze theorem.
    • To understand why $\displaystyle \lim_{x\to 0}\dfrac{\sin(𝒙)}{𝒙}=1$, we can use a geometric approach involving areas in the unit circle. Here's a step-by-step explanation:
      • 1. Unit Circle Setup:
        • Consider a unit circle centered at the origin.
        • Let $𝒙$ be an angle near $0$ on the positive $𝒙$-axis.
      • 2. First Triangle; Two possible interpretations:
        • Option 1: Right triangle
          • Base $=\cos(𝒙)$, height $=\sin(𝒙)$.
          • Area $=\dfrac{1}{2}\cdot \cos(𝒙)\cdot \sin(𝒙)$.
          • It is a right triangle formed by dropping a perpendicular from the circle to the $x$-axis.
        • Option 2: General triangle (more accurate for this limit discussion)
          • Base $=1$ (the radius of the unit circle).
          • Height $=\sin(𝒙)$.
          • Area $=\dfrac{1}{2}\cdot 1\cdot \sin(𝒙)=\dfrac{1}{2}\sin(𝒙)$.
          • This is a scalene triangle (three unequal sides), not necessarily a right triangle.
        • 📝N͟O͟T͟E͟: For the purpose of proving the limit $\displaystyle \lim_{x\to 0}\dfrac{\sin(𝒙)}{𝒙}=1$, we usually consider the triangle where the base is $1$ and the height is $\sin(𝒙)$, leading to the simpler expression $\text{Area}=\dfrac{1}{2}\sin(𝒙)$.
      • 3. Sector Area (Area $=\dfrac{1}{2}\cdot 𝒙$):
        • The sector is formed by the angle $𝒙$ in the unit circle.
        • The area formula for a sector is $\dfrac{1}{2}\,r^{2}\theta$. Since $r=1$, it simplifies to $\dfrac{1}{2}\,𝒙$.
      • 4. Larger Triangle Involving the Tangent Line:
        • Extend the tangent line from the point where the terminal side meets the circle.
        • his creates a larger triangle whose height corresponds to $\tan(𝒙)$.
      • 5. Setting Up Inequalities:
        • Comparing areas:
          • $\dfrac{1}{2}\sin(𝒙)<\dfrac{1}{2}\,𝒙<\dfrac{1}{2}\tan(𝒙)$
      • 6. Simplifying the Inequalities:
        • Multiply through by $2$:
          • $\sin(𝒙)<𝒙<\tan(𝒙)$< /li>
        • Divide all parts by $\sin(𝒙)$:
          • $1<\dfrac{𝒙}{\sin(𝒙)}<\dfrac{1}{\cos(𝒙)}$< /li>
        • By taking the reciprocal of all positive terms, the inequality reverses its direction:
          • $1<\dfrac{𝒙}{\sin(𝒙)}<\dfrac{1}{\cos(𝒙)}\ \Longrightarrow\ \cos(𝒙)<\dfrac{\sin(𝒙)}{𝒙}<1$
      • 7. Applying the Squeeze Theorem:
        • As $𝒙\to 0$, $\cos(𝒙)\to 1$, so $1/\cos(𝒙)\to 1$.
        • Thus, $𝒙/\sin(𝒙)$ is squeezed between $1$ and something approaching $1$.
        • By the Squeeze Theorem: $\displaystyle \lim_{x\to 0}\dfrac{\sin(𝒙)}{𝒙}=1$
    • 📝N͟O͟T͟E͟: In the proof using the Squeeze Theorem for $\displaystyle \lim_{x\to 0}\big(\sin(𝒙)/𝒙\big)=1$, it is correct to use the symbol “$<$” instead of “$\le$”. The reason is:
      • We are comparing areas or values strictly, not saying they are equal.
      • For very small $𝒙$ near $0$ (but not exactly $0$), we have strict inequalities.
      • The Squeeze Theorem only requires the function to be trapped between two functions, not necessarily touching them.
      • Thus, using “$<$” is perfectly fine and appropriate.
  • 1:55:00. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{1-\cos 𝒙}{𝒙}$ – [📷image]
    • $\displaystyle \lim_{x\to 0}\dfrac{1-\cos 𝒙}{𝒙}$ is calculated using the technique of multiplying by the conjugate.
    • The trigonometric identity $1-\cos^{2} 𝒙=\sin^{2} 𝒙$ is used to simplify the expression after multiplying by the conjugate.
    • Multiply and divide by the conjugate of the numerator $(1+\cos(𝒙))$:
      • $\displaystyle \lim_{x\to 0}\dfrac{(1-\cos(𝒙))(1+\cos(𝒙))}{𝒙\,(1+\cos(𝒙))}$
      • $\displaystyle \lim_{x\to 0}\dfrac{1-\cos^{2}(𝒙)}{𝒙\,(1+\cos(𝒙))}$
    • Use the identity: $1-\cos^{2}(𝒙)=\sin^{2}(𝒙)$
      • $\displaystyle \lim_{x\to 0}\dfrac{\sin^{2}(𝒙)}{𝒙\,(1+\cos(𝒙))}$
    • Split into two factors:
      • $\displaystyle \left(\lim_{x\to 0}\dfrac{\sin(𝒙)}{𝒙}\right)\cdot \left(\lim_{x\to 0}\dfrac{\sin(𝒙)}{1+\cos(𝒙)}\right)$
    • Apply known limits:
      • $\displaystyle \lim_{x\to 0}\dfrac{\sin(𝒙)}{𝒙}=1,\quad \lim_{x\to 0}\dfrac{\sin(𝒙)}{1+\cos(𝒙)}=\dfrac{0}{2}=0$
    • $\displaystyle \lim_{x\to 0}\dfrac{1-\cos 𝒙}{𝒙}=0$
  • 2:02:55. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{\tan 𝒙}{𝒙}$ – [📷image]
    • $\tan 𝒙=\dfrac{\sin 𝒙}{\cos 𝒙}$.
    • The expression is split into two limits:
      • $\displaystyle \lim_{x\to 0}\dfrac{\sin 𝒙}{𝒙}$ and $\displaystyle \lim_{x\to 0}\dfrac{1}{\cos 𝒙}$.
    • Rewrite tangent as sine over cosine:
      • $\displaystyle \lim_{x\to 0}\dfrac{\sin(𝒙)/\cos(𝒙)}{𝒙}$
    • Simplify the complex fraction:
      • $\displaystyle \lim_{x\to 0}\dfrac{\sin(𝒙)}{\cos(𝒙)}\cdot \dfrac{1}{𝒙} =\left(\lim_{x\to 0}\dfrac{\sin(𝒙)}{𝒙}\right) \cdot \left(\lim_{x\to 0}\dfrac{1}{\cos(𝒙)}\right) =1\cdot 1$
    • $\displaystyle \lim_{x\to 0}\dfrac{\tan 𝒙}{𝒙}=1$

  Ｍａｎｉｐｕｌａｔｉｏｎ ｏｆ Ｔｒｉｇｏｎｏｍｅｔｒｉｃ Ｌｉｍｉｔｓ

  • Manipulation of Trigonometric Limits
    • Various examples of manipulating trigonometric limits to match known identities are presented.
    • Examples with $\sin 𝒙/𝒙$, $(1-\cos 𝒙)/𝒙$, and $\tan 𝒙/𝒙$ are included.
    • Techniques such as multiplying by a constant, splitting the limit into multiple limits, and using trigonometric identities are used.
    • "We transform the unknown into something known" using mathematical tools such as simplifications, identities, or properties.
  • 2:08:00. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{\sin(2𝒙)}{𝒙}$ – [📷image]
    • Multiply by $2/2$ to create a form we can use:
      • $\displaystyle \lim_{x\to 0}\dfrac{\sin(2𝒙)}{𝒙}\cdot \dfrac{2}{2}$
    • Rearrange the expression:
      • $\displaystyle \lim_{x\to 0}\dfrac{2\sin(2𝒙)}{2𝒙}$
    • Factor out the constant $2$:
      • $\displaystyle 2\cdot \lim_{x\to 0}\dfrac{\sin(2𝒙)}{2𝒙}$
    • Recognize the standard limit (with $u=2𝒙$):
      • $2\cdot \lim_{u\to 0}\dfrac{\sin u}{u}=2$
    • $\displaystyle \lim_{x\to 0}\dfrac{\sin(2𝒙)}{𝒙}=2$
  • 2:13:40. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{\sin(5𝒙)}{\sin(6𝒙)}$ – [📷image]
    • Multiply numerator and denominator by $1/𝒙$:
      • $\displaystyle \lim_{x\to 0}\dfrac{\sin(5𝒙)}{𝒙}\Big/\dfrac{\sin(6𝒙)}{𝒙}$
    • Multiply numerator and denominator by constants:
      • $\displaystyle \lim_{x\to 0}\left(\dfrac{\sin(5𝒙)}{𝒙}\cdot \dfrac{5}{5}\right)\Big/\left(\dfrac{\sin(6𝒙)}{𝒙}\cdot \dfrac{6}{6}\right)$
    • Rearrange the expression:
      • $\displaystyle \dfrac{5}{6}\cdot \lim_{x\to 0}\dfrac{\sin(5𝒙)}{5𝒙}\Big/\lim_{x\to 0}\dfrac{\sin(6𝒙)}{6𝒙}$
    • Apply the standard limit identity ($u=5𝒙,\ 6𝒙$):
      • $\displaystyle \lim_{x\to 0}\dfrac{\sin u}{u}=1$
    • So: $\dfrac{5}{6}\cdot \dfrac{1}{1}=\dfrac{5}{6}$
    • Therefore: $\displaystyle \lim_{x\to 0}\dfrac{\sin(5𝒙)}{\sin(6𝒙)}=\dfrac{5}{6}$
  • 2:20:25. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{\sin(𝒙^{2})}{𝒙}$ – [📷image]
    • Rewrite as a product:
      • $\displaystyle \lim_{x\to 0}\left(\dfrac{\sin(𝒙^{2})}{𝒙}\right)\cdot \dfrac{𝒙}{𝒙}$
    • Rearranged:
      • $\displaystyle \lim_{x\to 0}\, 𝒙\cdot \dfrac{\sin(𝒙^{2})}{𝒙^{2}}$
    • Split the limit:
      • $\displaystyle \big(\lim_{x\to 0} 𝒙\big)\cdot \big(\lim_{x\to 0}\dfrac{\sin(𝒙^{2})}{𝒙^{2}}\big)$
    • Use the standard limit identity:
      • $\displaystyle \lim_{u\to 0}\dfrac{\sin u}{u}=1 \quad (u=𝒙^{2})$
    • Therefore:
      • $0\cdot 1 = 0$
    • $\displaystyle \lim_{x\to 0}\dfrac{\sin(𝒙^{2})}{𝒙}=0$
  • 2:24:20. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{\sin^{2}(𝒙)}{𝒙}$ – [📷image]
    • Use the identity:
      • $\sin^{2}(𝒙)=\sin(𝒙)\cdot \sin(𝒙)$
    • Rewrite the expression:
      • $\displaystyle \lim_{x\to 0}\left(\dfrac{\sin(𝒙)}{𝒙}\right)\cdot \sin(𝒙)$
    • Split the limit:
      • $\displaystyle \left(\lim_{x\to 0}\dfrac{\sin(𝒙)}{𝒙}\right)\cdot \left(\lim_{x\to 0}\sin(𝒙)\right)$
    • Apply known limits:
      • $1 \cdot 0 = 0$
    • $\displaystyle \lim_{x\to 0}\dfrac{\sin^{2}(𝒙)}{𝒙}=0$
  • 2:25:45. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \sin\!\left(\dfrac{1}{𝒙}\right)$ (Graphical approach) – [📷image]
    • $\displaystyle \lim_{x\to 0}\sin\!\left(\dfrac{1}{𝒙}\right)=\text{D.N.E}$
    • Graph behavior:
      • As $x$ approaches $0$, $1/x$ increases rapidly in magnitude.
      • This makes the sine function oscillate faster and faster.
    • Infinite waves:
      • The graph oscillates infinitely between $-1$ and $1$ as $x\to 0$.
      • It doesn't "settle" on any value — it doesn’t flatten or approach a specific number.
    • No fixed value:
      • For the limit to exist, $\sin(1/x)$ would need to get closer to a single number.
      • But instead, it keeps jumping around with no stabilization.
      • There are infinitely many peaks and valleys, getting tighter near $x=0$.
    • Conclusion:
      • Since the function doesn't approach any one value as $x\to 0$, the limit does not exist.
      • The graph shows a dense “band” of oscillation between $-1$ and $1$ near $x=0$.
  • 2:25:45. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ x\cdot \sin\!\left(\dfrac{1}{𝒙}\right)$ – [📷image]
    • The correct approach is to apply the Squeeze Theorem:
    • We know that:
      • $-1\le \sin(1/𝒙)\le 1$
    • Multiplying all parts by $𝒙$ (which tends to $0$), we get:
      • $-|𝒙|\le 𝒙\cdot \sin(1/𝒙)\le |𝒙|$
      • 📝N͟O͟T͟E͟:
        • we know that:
          • $-1\le \sin(1/𝒙)\le 1$
        • When multiplying by $𝒙$, the direction of the inequalities depends on the sign of $𝒙$:
          • If $𝒙>0$, the inequality remains the same.
          • If $𝒙<0$, the inequality reverses direction.
        • To avoid this problem, we use the absolute value on the bounds of the inequality, as follows:
          • $-|𝒙|\le 𝒙\,\sin(1/𝒙)\le |𝒙|$
    • We introduce the absolute value to keep the direction of the inequality consistent, since multiplying by $𝒙$ would reverse the inequality when $𝒙<0$. Using $|𝒙|$ ensures the inequality remains $-|𝒙|\le 𝒙\cdot \sin(1/𝒙)\le |𝒙|$ for both positive and negative $𝒙$.
    • Apply limits:
      • $\displaystyle \lim_{x\to 0}\big(-|𝒙|\big)\ \le\ \lim_{x\to 0}\big(𝒙\cdot \sin(1/𝒙)\big)\ \le\ \lim_{x\to 0}|𝒙|$
    • And since:
      • $\displaystyle \lim_{x\to 0}|𝒙|=0\quad \Rightarrow\quad 0\le \lim_{x\to 0}\big(𝒙\cdot \sin(1/𝒙)\big)\le 0$
    • By the Squeeze Theorem, it follows that:
      • $\displaystyle \lim_{x\to 0}\, 𝒙\cdot \sin\!\left(\dfrac{1}{𝒙}\right)=0$
    • ⚠📝 NOTE: At first glance, the limit:
      • $\displaystyle \lim_{x\to 0} 𝒙 \cdot \sin\!\left(\dfrac{1}{𝒙}\right)$
      • might look like it fits the identity:
        • $\displaystyle \lim_{u\to 0} \dfrac{\sin u}{u}=1$
      • But that identity 𝐨𝐧𝐥𝐲 𝐚𝐩𝐩𝐥𝐢𝐞𝐬 when the argument of the sine and the denominator both approach $0$ at the same time. Now, if we rewrite:
        • $\displaystyle 𝒙 \cdot \sin\!\left(\dfrac{1}{𝒙}\right)=\dfrac{\sin(1/𝒙)}{1/𝒙}$
          • Here, as $𝒙\to 0$:
            • $1/𝒙\to \infty$,
            • and $\sin(1/𝒙)$ keeps oscillating between $-1$ and $1$ with no limit.
      • So the expression $\sin(1/𝒙)/(1/𝒙)$ doesn't approach $1$, and we 𝐜𝐚𝐧𝐧𝐨𝐭 apply the identity.
    • 2:35:33. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{2-\cos(3𝒙)-\cos(4𝒙)}{𝒙}$ – [📷image]
      • Rewrite the expression:
        • $\displaystyle \lim_{x\to 0}\dfrac{2-\cos(3𝒙)-\cos(4𝒙)}{𝒙}$
      • Use the identity:
        • $2-\cos(3𝒙)-\cos(4𝒙)=\big(1-\cos(3𝒙)\big)+\big(1-\cos(4𝒙)\big)$
      • Split into two separate limits:
        • $\displaystyle \lim_{x\to 0}\dfrac{1-\cos(3𝒙)}{𝒙}\ +\ \lim_{x\to 0}\dfrac{1-\cos(4𝒙)}{𝒙}$
      • Multiply and divide to match the standard form:
        • $\displaystyle \lim_{x\to 0}\dfrac{1-\cos(3𝒙)}{3𝒙}\cdot 3\ +\ \lim_{x\to 0}\dfrac{1-\cos(4𝒙)}{4𝒙}\cdot 4$
      • Apply the known limit:
        • $\displaystyle \lim_{u\to 0}\dfrac{1-\cos u}{u}=0$
      • Evaluate:
        • $3\cdot 0+4\cdot 0=0$
      • Final result:
        • $\displaystyle \lim_{x\to 0}\dfrac{2-\cos(3𝒙)-\cos(4𝒙)}{𝒙}=0$
    • 2:40:45. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{𝒙^{2}-3\sin(𝒙)}{𝒙}$ – [📷image]
      • Rewrite the expression:
        • $\displaystyle \lim_{x\to 0}\dfrac{𝒙^{2}-3\sin(𝒙)}{𝒙}$
      • Split the terms:
        • $\displaystyle \lim_{x\to 0}\left(\dfrac{𝒙^{2}}{𝒙}\right)-3\cdot \left(\dfrac{\sin(𝒙)}{𝒙}\right)$
      • Simplify each part:
        • $\displaystyle \lim_{x\to 0} 𝒙-3\cdot \left(\dfrac{\sin(𝒙)}{𝒙}\right)$
      • Apply limits:
        • $\displaystyle \lim_{x\to 0} 𝒙=0$
        • $\displaystyle \lim_{x\to 0}\dfrac{\sin(𝒙)}{𝒙}=1$
      • Compute:
        • $0-3\cdot 1=-3$
      • Final result:
        • $\displaystyle \lim_{x\to 0}\dfrac{𝒙^{2}-3\sin(𝒙)}{𝒙}=-3$
    • 2:42:24. 🧩 Evaluation of $\displaystyle \lim_{t\to 0}\ \dfrac{t^{2}}{1-\cos^{2}(t)}$ – [📷image]
      • Use the trigonometric identity:
        • $1-\cos^{2}(t)=\sin^{2}(t)$
      • Rewrite the limit:
        • $\displaystyle \lim_{t\to 0}\dfrac{t^{2}}{1-\cos^{2}(t)}=\lim_{t\to 0}\dfrac{t^{2}}{\sin^{2}(t)}=\lim_{t\to 0}\left(\dfrac{t}{\sin t}\right)^{2}$
      • Recognize the limit structure:
        • $\displaystyle \lim_{t\to 0}\dfrac{\sin T}{T}=1$
        • Then: $\displaystyle \lim_{t\to 0}\dfrac{t}{\sin t}=\left(\lim_{t\to 0}\dfrac{\sin t}{t}\right)^{-1}=1^{-1}=1$
      • Apply the limit:
        • $\big(\lim_{t\to 0} t/\sin t\big)^{2}=1^{2}=1$
      • Final result: $1$
    • 2:50:00. 🧩 Evaluation of $\displaystyle \lim_{x\to 0}\ \dfrac{𝒙}{\cos(\pi/2-𝒙)}$ – [📷image]
      • Use the co-function identity:
        • $\cos(\pi/2-𝒙)=\sin(𝒙)$
        • 📝 N͟O͟T͟E͟: Two angles are complementary if they add up to $90^\circ$:
          • $\sin(\text{angle})\ \leftrightarrow\ \cos(\text{complement of angle})$
      • Replace the expression using the identity:
        • $\displaystyle \lim_{x\to 0}\dfrac{𝒙}{\sin(𝒙)}$
      • Recognize the standard limit:
        • $\displaystyle \lim_{x\to 0}\dfrac{𝒙}{\sin(𝒙)}=1$
    • 2:53:40. 🧩 Evaluation of $\displaystyle \lim_{\theta\to 0}\ \dfrac{\theta^{2}}{\,1-\cos(\theta)\,}$ – [📷image]
      • Multiply by $1$ in the form of $\dfrac{1+\cos(\theta)}{1+\cos(\theta)}$ to simplify:
        • $\displaystyle \lim_{\theta\to 0}\dfrac{\theta^{2}\big(1+\cos(\theta)\big)}{1-\cos^{2}(\theta)}$
      • Use the identity: $1-\cos^{2}(\theta)=\sin^{2}(\theta)$
        • $\displaystyle \lim_{\theta\to 0}\dfrac{\theta^{2}\big(1+\cos(\theta)\big)}{\sin^{2}(\theta)}$
      • Separate into two limits:
        • $\displaystyle \left(\lim_{\theta\to 0}\dfrac{\theta^{2}}{\sin^{2}(\theta)}\right)\cdot \left(\lim_{\theta\to 0}1+\cos(\theta)\right)$
      • Use the identity:
        • $\displaystyle \lim_{\theta\to 0}\dfrac{\theta}{\sin(\theta)}=1$, so:
          • $\displaystyle \lim_{\theta\to 0}\left(\dfrac{\theta}{\sin(\theta)}\right)^{2}=1^{2}=1$
      • Evaluate the limits:
        • $1\cdot \big(1+\cos(0)\big)=1\cdot (1+1)=2$