2.6 Limits at Infinity; Horizontal Asymptotes — Study Notes '' '

Problem 1

Goal: Evaluate the limits as x → ∞ and x → −∞ for the rational function and identify any horizontal asymptote.
Expression:
f(x)=\frac{2x^3+x^2-1}{5x^3-7x+2}
Approach:
- For large |x|, compare leading terms. Divide numerator and denominator by $x^3$ or read off leading coefficients.
Results:
- \lim_{x\to\infty}\frac{2x^3+x^2-1}{5x^3-7x+2}=\frac{2}{5}
- \lim_{x\to-\infty}\frac{2x^3+x^2-1}{5x^3-7x+2}=\frac{2}{5}
Horizontal asymptote:
- $y=\tfrac{2}{5}$ since the limits at both ends are finite and equal to the same value.
Quick method:
- Divide numerator and denominator by $x^3$:
  \frac{2+\tfrac{1}{x}-\tfrac{1}{x^3}}{5-\tfrac{7}{x^2}+\tfrac{2}{x^3}} \to \frac{2}{5}.

Expression:
g(x)=\frac{4x^2-2x+3}{5-3x}
Limit as $x\to\infty$:
- Leading behavior: numerator ~ $4x^2$, denominator ~ $-3x$; ratio ~ $-(4/3)x \to -\infty$.
- \lim_{x\to\infty} g(x)= -\infty
Horizontal asymptote:
- None (no finite horizontal limit as $x \to \infty$). Note: there is no horizontal line $y=c$ that $f(x)$ approaches.
Optional note on oblique asymptote:
- Since degrees differ by 1, there is an oblique (slant) asymptote. Long division gives
  \frac{4x^2-2x+3}{5-3x} = -\frac{4}{3}x - \frac{14}{9} + \frac{97}{9(5-3x)}.
- Oblique asymptote: $y=-\frac{4}{3}x-\frac{14}{9}$.

Expression:
h(x)=\frac{x^3(x-8)^2}{x^4(2x^2+9)}
Simplify:
h(x)=\frac{(x-8)^2}{x(2x^2+9)}
Limit as $x\to -\infty$:
- Numerator ~ $x^2$, denominator ~ $2x^3$; ratio ~ $\frac{x^2}{2x^3}=\frac{1}{2x}$ → 0.
Result:
- \lim_{x\to-\infty} h(x)=0

Expression:
k(x)=\frac{\sqrt{9x^2-12}}{5x+2}
Big-$|x|$ behavior:
- $\sqrt{9x^2-12}=3|x|\sqrt{1-\frac{12}{9x^2}}$.
For $x\to -\infty$ we have $|x|=-x$.
Simplify:
k(x)=\frac{3|x|\sqrt{1-\frac{12}{9x^2}}}{5x+2}=\frac{3(-x)\sqrt{1-\frac{4}{3x^2}}}{x\left(5+\frac{2}{x}\right)}=\frac{-3\sqrt{1-\frac{4}{3x^2}}}{5+\frac{2}{x}}.
Take the limit as $x\to-\infty$:
- Numerator tends to $-3$, denominator tends to $5$, so
- \lim_{x\to-\infty} k(x) = -\frac{3}{5}.

Expression:
m(x)=\frac{\sqrt{x^2-x}-x}{2}
Strategy: rationalize using conjugate.
Compute:
\sqrt{x^2-x}-x=\frac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=\frac{-x}{\sqrt{x^2-x}+x}.
Therefore
m(x)=\frac{-x}{2(\sqrt{x^2-x}+x)}.
Limit as $x\to\infty$:
- Write ratio in terms of $\sqrt{x^2-x}/x = \sqrt{1-\frac{1}{x}} \to 1$.
- Then
  \frac{x}{\sqrt{x^2-x}+x} = \frac{1}{\frac{\sqrt{x^2-x}}{x}+1} \to \frac{1}{1+1}=\frac{1}{2}.
- Hence
  \lim_{x\to\infty} m(x) = \frac{-1}{2}\cdot\frac{1}{2} = -\frac{1}{4}.

Expression:
p(x)=\frac{4+3e^{x}}{3+e^{-x}}
Limit as $x\to-\infty$:
- $e^{x}\to 0$, $e^{-x}\to \infty$.
- Numerator $\to 4$, denominator $\to \infty$; overall goes to 0.
Result:
- \lim_{x\to-\infty} p(x)=0.

Expression:
q(x)=\frac{3e^{2x}+e^{-3x}}{e^{2x}-4e^{-3x}}
Factor $e^{2x}$:
q(x)=\frac{e^{2x}(3+e^{-5x})}{e^{2x}(1-4e^{-5x})} = \frac{3+e^{-5x}}{1-4e^{-5x}}.
Limit as $x\to\infty$:
- $e^{-5x}\to 0$; hence $q(x)\to 3/1 = 3$.
Result:
- \lim_{x\to\infty} q(x)=3.

Expression:
t(x)=\ln\left(\frac{x+4}{3x+7}\right),\quad x\to\infty.
Limit of the inside ratio:
- \frac{x+4}{3x+7} = \frac{1 + 4/x}{3 + 7/x} \to \frac{1}{3}.
Therefore
- \lim_{x\to\infty} t(x) = \ln\left(\frac{1}{3}\right) = -\ln 3.

Expression:
u(x)=\ln(x^2-9) - \ln(x^3+5) = \ln\left(\frac{x^2-9}{x^3+5}\right).
Limit as $x\to\infty$:
- Inside ratio: $(x^2-9)/(x^3+5) \sim x^2/x^3 = 1/x \to 0^+$.
- Hence $\ln(\text{something that goes to }0^+)$ tends to $-\infty$.
Result:
- \lim_{x\to\infty} [\ln(x^2-9) - \ln(x^3+5)] = -\infty.

Expression:
v(x)=\arctan\left(\frac{x^2-9}{x^2+3x-5}\right),\quad x\to\infty.
Inside ratio:
- Degree both numerator and denominator is 2; leading terms give $\frac{x^2}{x^2}=1$, so the ratio $\to 1$ as $x\to\infty$.
Therefore
- \lim_{x\to\infty} v(x) = \arctan(1) = \frac{\pi}{4}.

Expression:
w(x)=\arctan\left(\frac{x-5}{x-3}\right),\quad x\to 3^+.
As $x\to 3^+$: numerator $(x-5)\to -2$, denominator $(x-3)\to 0^+$, so the fraction (\to -\infty).
Therefore
- \lim_{x\to 3^+} w(x) = \arctan(-\infty) = -\frac{\pi}{2}.

Expression:
y(x)=\arctan(\ln x),\quad x\to\infty.
Behavior:
- $\ln x \to \infty$, so $\arctan(\ln x) \to \arctan(\infty) = \frac{\pi}{2}$.
Result:
- \lim_{x\to\infty} \arctan(\ln x) = \frac{\pi}{2}.

Horizontal asymptotes for rational functions: compare leading terms; if degrees equal, limit is ratio of leading coefficients; if finite, horizontal asymptote exists with that y-value.
When degrees differ by 1 or more, horizontal asymptotes may not exist; oblique/slant asymptotes can occur (found via polynomial division).
Limits involving radicals with $\sqrt{x^2}$ require using $\sqrt{x^2}=|x|$; for $x\to-\infty$, $|x|=-x$.
Exponential functions: factor out dominant exponential to simplify limits; e.g., divide numerator and denominator by $e^{2x}$ when $x\to\infty$.
Logs: use log properties to simplify differences and quotients; track leading growth to determine finite limits or signs of infinity.
Vertical asymptotes: occur when denominator approaches 0 and numerator does not approach 0; examine one-sided limits to determine the direction of infinity.
Conjugate trick for limits of the form (\sqrt{ax^2+bx+c}-dx) as (x\to\infty): multiply by the conjugate to remove the square root and extract the finite limit.
All numerical results (limits) in this set are given in LaTeX form above where appropriate.

Problem 1: \lim{x\to\infty}=\frac{2}{5},\quad \lim{x\to-\infty}=\frac{2}{5},\quad \text{horizontal asymptote } y=\frac{2}{5}.
Problem 2: \lim_{x\to\infty}=-\infty;\quad \text{no horizontal asymptote; oblique asymptote } y=-\frac{4}{3}x-\frac{14}{9}.
Problem 3: \lim_{x\to-\infty}=0.
Problem 4: \lim_{x\to-\infty}= -\frac{3}{5}.
Problem 5: \lim_{x\to\infty}= -\frac{1}{4}.
Problem 6: \lim_{x\to-\infty}=0.
Problem 7: \lim_{x\to\infty}=3.
Problem 8: \lim_{x\to-3^+}=+\infty.
Problem 9: \lim_{x\to-3^-}=-\infty.
Problem 10: \lim_{x\to\infty}= -\ln 3.
Problem 11: \lim_{x\to\infty}= -\infty.
Problem 12: \lim_{x\to\infty}=\frac{\pi}{4}.
Problem 13: \lim_{x\to 3^+}= -\frac{\pi}{2}.
Problem 14: \lim_{x\to\infty}=\frac{\pi}{2}.