Ｃａｌｃｕｌｕｓ　１　ｌｅｃｔｕｒｅ　３．５

　Lｉｍｉｔｓ　Oｆ　Fｕｎｃｔｉｏｎｓ　ａｔ　Iｎｆｉｎｉｔｙ

Ｒｅｖｉｅｗ　ｏｆ　ｖｅｒｔｉｃａｌ　ａｓｙｍｐｔｏｔｅｓ　ａｎｄ　ｒｅｍｏｖａｂｌｅ　ｄｉｓｃｏｎｔｉｎｕｉｔｉｅｓ　

• [0:00]. Reminder of the definition of vertical asymptote (VA): $\lim_{x\to a} 𝒇(𝒙)=\pm\infty$ [📷image]
  • Two cases for a vertical asymptote:
    • $\lim_{x\to a^-} 𝒇(𝒙)=\lim_{x\to a^+} 𝒇(𝒙)=\pm\infty$   # The limit exists (both one-sided limits agree)
    • $\lim_{x\to a^-} 𝒇(𝒙)=\infty$ and $\lim_{x\to a^+} 𝒇(𝒙)=-\infty$  # The limit does not exist (one-sided limits disagree)
  • to find discontinuities denominator equals zero.
    • If a factor cannot be canceled, the discontinuity is a vertical asymptote.
    • If a factor can be canceled, it is a removable discontinuity, not a vertical asymptote.
• [3:55]. Difference between removable discontinuities (holes) and VA. [📷image]
  • ʜ̳ᴏ̳ʟ̳ᴇ̳ꜱ̳: removable discontinuities where only one point is missing in the function.
    • Occur when the numerator and denominator are zero at the same time, allowing cancellation of factors.
    • One single point is missing from the function.
  • ᴠ̳ᴀ̳: non-removable discontinuities.
    • Occur when discontinuities cannot be canceled.
• [6:40]. 🧩 Example –: $𝒇(𝒙)=\dfrac{𝒙}{(𝒙+3)(𝒙-1)}$ [📷image]
  • Vertical asymptotes at $𝒙=1$ and $𝒙=-3$ → non-cancelable factors in the denominator.
  • Critical sign-change values are found by:
    • Setting the denominator equal to zero → to identify discontinuities (vertical asymptotes).
    • Setting the numerator equal to zero → to identify values where the function crosses the $𝒙$-axis.
      • These are not discontinuities, but they must be included in the sign analysis since they may cause a sign change in the rational expression.
        • In this case, $𝒙=0$ makes the numerator zero → the function is zero at this point and may change sign around it.
  • General rule for analyzing a rational function $𝒇(𝒙)=\dfrac{𝒫(𝒙)}{𝒬(𝒙)}$:
    • Fully factor both $𝒫(𝒙)$ and $𝒬(𝒙)$, and simplify the expression if possible.
    • Identify points of discontinuity from $𝒬(𝒙)=0$:
      • If a factor in the denominator does not cancel with the numerator, its zero corresponds to a vertical asymptote (VA).
        • These values are not in the domain and divide the number line into intervals.
        • They are essential in the sign chart, as they may trigger sign changes in $𝒇(𝒙)$.
      • If a factor in the denominator does cancel with one in the numerator, the corresponding $𝒙$-value results in a hole (removable discontinuity).
        • The function is undefined at this point, but the behavior around it may be continuous.
        • Holes do not cause sign changes like VAs do, but must still be noted in the analysis.
    • Set $𝒫(𝒙)=0$ → determine x-intercepts (zeros of the numerator).
      • Include them in the sign analysis only if they are in the domain, as they may indicate a sign change in the rational expression.
  • [8:14]. Using sign analysis to determine behavior around VAs:
    • Sign chart:
    • Limits at critical values:
      • Near $𝒙=-3$ (vertical asymptote):
        • $\lim_{x\to-3^-} 𝒇(𝒙)=-\infty$ # denominator $\to0^+$, numerator $<0$ ⇒ $-\infty$
        • $\lim_{x\to-3^+} 𝒇(𝒙)=+\infty$ # denominator $\to0^-$, numerator $<0$ ⇒ $+\infty$
        • Limit does not exist.
      • At $𝒙=0$ (zero of the numerator):
        • $\lim_{x\to0} 𝒇(𝒙)=0$ # numerator $\to0$, denominator $\neq0$ ⇒ crosses the x-axis.
      • Near $𝒙=1$ (vertical asymptote):
        • $\lim_{x\to1^-} 𝒇(𝒙)=-\infty$ # denominator $\to0^-$, numerator $>0$ ⇒ $-\infty$
        • $\lim_{x\to1^+} 𝒇(𝒙)=+\infty$ # denominator $\to0^+$, numerator $>0$ ⇒ $+\infty$
        • Limit does not exist.
• [11:55]. 🧩 Example –: $\lim_{t\to1} \dfrac{t^3}{(t^2-1)^2}$ [📷image]
  • Discontinuities:
    • Discontinuities occur where the denominator equals zero.
    • For this problem, they happen at $t=\pm1$ because $(t^2-1)^2=0$.
    • These are vertical asymptotes as they cannot be removed by factoring (non-removable discontinuities).
  • Focus on the limit:
    • Even though there are two discontinuities ($t=\pm1$), only $t=1$ matters since the limit is evaluated near this value.
    • $t=-1$ can be ignored for this problem.
  • Importance of the numerator:
    • It is crucial for calculating the limit at vertical asymptotes (VA), as these points determine where the function may change its sign in the adjacent intervals.
    • Check where the numerator ($t^3=0$) equals zero, which happens at $t=0$.
    • $t=0$ is not an asymptote, but it is useful to divide intervals and check for sign changes.
  • Strategy:
    • Divide the interval into regions around $t=0$ and $t=1$.
    • Evaluate specific points in each region (e.g., $t=0.5,\; t=2$) to understand the function's behavior.
    • Sign chart:
      • Near $x=1$ (vertical asymptote):
        • $\lim_{x\to1^-} 𝒇(𝒙)=+\infty$ # denominator $\to0^-$, numerator $>0$ ⇒ $+\infty$
        • $\lim_{x\to1^+} 𝒇(𝒙)=+\infty$ # denominator $\to0^+$, numerator $>0$ ⇒ $+\infty$
        • Limit exist: $\lim_{x\to1} 𝒇(𝒙)=+\infty$

Ｌｉｍｉｔｓ　ａｔ　ｉｎｆｉｎｉｔｙ

• [16:30]. Question: What happens to a function as $𝒙$ approaches positive or negative infinity? [📷image]
  • Exploring the behavior of $1/𝒙$ as $𝒙$ increases:
    • As $𝒙$ increases, $1/𝒙$ approaches zero.
      • $\lim_{x\to\infty} \dfrac{1}{x}=0$
      • $\lim_{x\to-\infty} \dfrac{1}{x}=0$
• [20:25]. Definition of limit at infinity :
  • If $𝒇(𝒙)$ approaches a number as $𝒙$ approaches infinity, the limit exists.
• [21:07]. Relation between limits at infinity and horizontal asymptotes (HA):
  • HA represent the value a function approaches as $𝒙$ goes to positive or negative infinity.
• [23:00]. The rules for calculating limits as $𝒙\to+\infty$ or $𝒙\to-\infty$ [📷image] are the same as before. Therefore, the same rules and techniques apply in both cases.
  • $\lim_{x\to\infty} \dfrac{1}{x^n}=0$ for any $n>0$ :
    • Any constant divided by a variable raised to a power approaching infinity will approach zero.
• [26:10]. Behavior of polynomials at infinity [📷image]
  • Polynomials do not approach a specific number at infinity but tend to positive or negative infinity.
  • The behavior of a polynomial at infinity is determined by the highest power term.
    • This is because, as $𝒙$ approaches infinity, the lower exponent terms become insignificant compared to the dominant term.
  • 🧩 Example –: $\lim_{x\to-\infty} \left(-3x^3-2x^2-x+9\right)=\lim_{x\to-\infty} (-3x^3)=\infty$

Ｃａｌｃｕｌａｔｉｎｇ　ｌｉｍｉｔｓ　ａｔ　ｉｎｆｉｎｉｔｙ

• [36:27]. Introduction.
• [38:15]. 🧩 Example –: $\lim_{x\to\infty} \dfrac{5x-2}{3x+9}$ [📷image]
  • Key idea: divide each term by the highest power of $𝒙$ in the denominator.
  • Terms with $𝒙$ in the denominator approach zero as $𝒙$ approaches infinity.
  • $\lim_{x\to\infty} \dfrac{5-2/x}{3+9/x}=\dfrac{5}{3}$
  • Observation: the limit at infinity of a rational function where the powers of $𝒙$ in the numerator and denominator are equal is the ratio of the leading coefficients.
• [44:28]. 🧩 Example –: $\lim_{x\to-\infty} \dfrac{5x^2-4x}{15x^3-3x}$ [📷image]
  • Divide all terms by the highest power of $x$ in the denominator ($x^3$):
    • $\lim_{x\to-\infty} \dfrac{5/x-4/x^2}{15-3/x^3}$
  • Analyze behavior as $x\to-\infty$:
    • $5/x\to0$, $4/x^2\to0$, $3/x^3\to0$
  • Substitute limits:
    • $\dfrac{0-0}{15-0}=0$ $\Rightarrow\ \lim_{x\to-\infty} \dfrac{5x^2-4x}{15x^3-3x}=0$
• [48:05]. 🧩 Example –: $\lim_{x\to-\infty} \dfrac{7x^3-2x^2+1}{9-2x}$ [📷image]
  • Divide all terms by the highest power of $x$ in the denominator ($x$):
    • $\lim_{x\to-\infty} \dfrac{7x^2-2x+1/x}{9/x-2}$
  • Analyze behavior as $x\to-\infty$:
    • $7x^2\to+\infty$, $-2x\to+\infty$, $1/x\to0$, $9/x\to0$, denominator $\to-2$
  • Substitute limits:
    • $\dfrac{+\infty}{-2}=-\infty$ $\Rightarrow\ \lim_{x\to-\infty} \dfrac{7x^3-2x^2+1}{9-2x}=-\infty$
• [52:15]. 🧩 Example –: $\lim_{x\to\infty} \sqrt[3]{\dfrac{2x^2-3}{x^2-5}}$ [📷image]
  • Divide inside by the highest power $x^2$:
    • $\sqrt[3]{\dfrac{2-3/x^2}{1-5/x^2}}$
  • Take the limit:
    • $\sqrt[3]{\dfrac{2}{1}}=\sqrt[3]{2}$
• [53:50]. 🧩 Example –: $\lim_{x\to-\infty} \dfrac{\sqrt{x^2+2}}{3x-6}$ [📷image-1] [📷image-2] [📷image-3]
  • Divide by $x$ (highest power in the denominator):
    • ➀ Applied Property: Multiplication of Terms with Fractional Powers: $a\cdot b=\sqrt{a^2b^2}$
    • 🧩 Example –: $\dfrac{1}{x}\cdot\sqrt{x^2+2}=\sqrt{\dfrac{x^2+2}{x^2}}$
    • ➁ Pay attention to the use of absolute value when simplifying square roots: $\sqrt{x^2}=|x|$
    • (i) Why is $x\neq\sqrt{x^2}$ in general?
      • If $x\ge0$: $\sqrt{x^2}=x$
      • If $x<0$: e.g., $x=-3\Rightarrow \sqrt{9}=3\neq -3$
    • (ii) Why is $|x|=\sqrt{x^2}$ always true?
      • Both yield the non-negative magnitude of $x$.
  • Step 1. Factor inside the root:
    • $x^2+2=x^2\,(1+2/x^2)\Rightarrow \sqrt{x^2+2}=|x|\sqrt{1+2/x^2}$
  • Step 2. Substitute:
    • $\dfrac{\sqrt{x^2+2}}{3x-6}=\dfrac{|x|\sqrt{1+2/x^2}}{3x-6}$
  • Step 3. Divide numerator/denominator by $x$:
    • $\dfrac{|x|/x\cdot\sqrt{1+2/x^2}}{3-6/x}$
  • Step 4. Limit as $x\to-\infty$:
    • $|x|/x\to-1$, $\sqrt{1+2/x^2}\to1$, $3-6/x\to3$
  • Conclusion:
    • $\lim_{x\to-\infty} \dfrac{\sqrt{x^2+2}}{3x-6}=\dfrac{-1\cdot1}{3}=-\dfrac{1}{3}$
• [1:06:53]. 🧩 Example –: $\lim_{x\to\infty}\big(\sqrt{x^4+2}-x^2\big)$ [📷image]
  • Rationalize with the conjugate.
  • Divide by $x^2$ (highest power relevant).
  • Step 1: Multiply by the conjugate
    • $\lim_{x\to\infty} \dfrac{(\sqrt{x^4+2}-x^2)(\sqrt{x^4+2}+x^2)}{\sqrt{x^4+2}+x^2}$
  • Step 2: $(a-b)(a+b)=a^2-b^2$
    • $\lim_{x\to\infty} \dfrac{x^4+2-x^4}{\sqrt{x^4+2}+x^2}=\lim_{x\to\infty} \dfrac{2}{\sqrt{x^4+2}+x^2}$
  • Step 3: Divide top/bottom by $x^2$
    • $\lim_{x\to\infty} \dfrac{\dfrac{2}{x^2}}{\sqrt{1+\dfrac{2}{x^4}}+1}$
  • Step 4: Take the limit
    • $\dfrac{0}{1+1}=0$
• [1:16:00]. 🧩 Example –: $\lim_{x\to\infty}\big(\sqrt{x^4+2x^2+x^2}-x^2\big)$ [📷image]
  • Rationalize with the conjugate.
  • Divide each term by $x^2$.
  • Step 1: Multiply by the conjugate
    • $\lim_{x\to\infty}\dfrac{(\sqrt{x^4+2x^2+x^2}-x^2)(\sqrt{x^4+2x^2+x^2}+x^2)}{\sqrt{x^4+2x^2+x^2}+x^2}$
  • Step 2: $(a-b)(a+b)=a^2-b^2$
    • $\lim_{x\to\infty}\dfrac{x^4+2x^2+x^2-x^4}{\sqrt{x^4+2x^2+x^2}+x^2}=\lim_{x\to\infty}\dfrac{2x^2}{\sqrt{x^4+2x^2+x^2}+x^2}$
  • Step 3: Divide by $x^2$
    • $\lim_{x\to\infty}\dfrac{2}{\sqrt{1+\dfrac{2}{x^2}+\dfrac{1}{x^2}}+1}$
  • Step 4: Take the limit
    • $\dfrac{2}{1+1}=1$
• [1:20:51]. 🧩 Example –: $\lim_{x\to\infty} \sqrt{7-x}$ [📷image]
  • Does not exist: D.N.E
  • The expression inside the square root becomes negative as $𝒙\to+\infty$.
• SUMMARY: INDETERMINATE FORMS
  When $x \to \infty$ or $x \to -\infty$ in limits