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 ) = 2 x 3 + x 2 − 1 5 x 3 − 7 x + 2 f(x)=\frac{2x^3+x^2-1}{5x^3-7x+2} f ( x ) = 5 x 3 − 7 x + 2 2 x 3 + x 2 − 1 Approach:For large |x|, compare leading terms. Divide numerator and denominator by $x^3$ or read off leading coefficients. Results:lim x → ∞ 2 x 3 + x 2 − 1 5 x 3 − 7 x + 2 = 2 5 \lim_{x\to\infty}\frac{2x^3+x^2-1}{5x^3-7x+2}=\frac{2}{5} lim x → ∞ 5 x 3 − 7 x + 2 2 x 3 + x 2 − 1 = 5 2 lim x → − ∞ 2 x 3 + x 2 − 1 5 x 3 − 7 x + 2 = 2 5 \lim_{x\to-\infty}\frac{2x^3+x^2-1}{5x^3-7x+2}=\frac{2}{5} lim x → − ∞ 5 x 3 − 7 x + 2 2 x 3 + x 2 − 1 = 5 2 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 $x^3$:2 + 1 x − 1 x 3 5 − 7 x 2 + 2 x 3 → 2 5 . \frac{2+\tfrac{1}{x}-\tfrac{1}{x^3}}{5-\tfrac{7}{x^2}+\tfrac{2}{x^3}} \to \frac{2}{5}. 5 − x 2 7 + x 3 2 2 + x 1 − x 3 1 → 5 2 . Problem 2 Expression:g ( x ) = 4 x 2 − 2 x + 3 5 − 3 x g(x)=\frac{4x^2-2x+3}{5-3x} g ( x ) = 5 − 3 x 4 x 2 − 2 x + 3 Limit as $x\to\infty$:Leading behavior: numerator ~ $4x^2$, denominator ~ $-3x$; ratio ~ $-(4/3)x \to -\infty$. lim x → ∞ g ( x ) = − ∞ \lim_{x\to\infty} g(x)= -\infty lim x → ∞ g ( x ) = − ∞ 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 gives4 x 2 − 2 x + 3 5 − 3 x = − 4 3 x − 14 9 + 97 9 ( 5 − 3 x ) . \frac{4x^2-2x+3}{5-3x} = -\frac{4}{3}x - \frac{14}{9} + \frac{97}{9(5-3x)}. 5 − 3 x 4 x 2 − 2 x + 3 = − 3 4 x − 9 14 + 9 ( 5 − 3 x ) 97 . Oblique asymptote: $y=-\frac{4}{3}x-\frac{14}{9}$. Problem 3 Expression:h ( x ) = x 3 ( x − 8 ) 2 x 4 ( 2 x 2 + 9 ) h(x)=\frac{x^3(x-8)^2}{x^4(2x^2+9)} h ( x ) = x 4 ( 2 x 2 + 9 ) x 3 ( x − 8 ) 2 Simplify:h ( x ) = ( x − 8 ) 2 x ( 2 x 2 + 9 ) h(x)=\frac{(x-8)^2}{x(2x^2+9)} h ( x ) = x ( 2 x 2 + 9 ) ( x − 8 ) 2 Limit as $x\to -\infty$:Numerator ~ $x^2$, denominator ~ $2x^3$; ratio ~ $\frac{x^2}{2x^3}=\frac{1}{2x}$ → 0. Result:lim x → − ∞ h ( x ) = 0 \lim_{x\to-\infty} h(x)=0 lim x → − ∞ h ( x ) = 0 Problem 4 Expression:k ( x ) = 9 x 2 − 12 5 x + 2 k(x)=\frac{\sqrt{9x^2-12}}{5x+2} k ( x ) = 5 x + 2 9 x 2 − 12 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 ) = 3 ∣ x ∣ 1 − 12 9 x 2 5 x + 2 = 3 ( − x ) 1 − 4 3 x 2 x ( 5 + 2 x ) = − 3 1 − 4 3 x 2 5 + 2 x . 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}}. k ( x ) = 5 x + 2 3∣ x ∣ 1 − 9 x 2 12 = x ( 5 + x 2 ) 3 ( − x ) 1 − 3 x 2 4 = 5 + x 2 − 3 1 − 3 x 2 4 . Take the limit as $x\to-\infty$:Numerator tends to $-3$, denominator tends to $5$, so lim x → − ∞ k ( x ) = − 3 5 . \lim_{x\to-\infty} k(x) = -\frac{3}{5}. lim x → − ∞ k ( x ) = − 5 3 . Problem 5 Expression:m ( x ) = x 2 − x − x 2 m(x)=\frac{\sqrt{x^2-x}-x}{2} m ( x ) = 2 x 2 − x − x Strategy: rationalize using conjugate. Compute:x 2 − x − x = ( x 2 − x ) − x 2 x 2 − x + x = − x x 2 − x + x . \sqrt{x^2-x}-x=\frac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=\frac{-x}{\sqrt{x^2-x}+x}. x 2 − x − x = x 2 − x + x ( x 2 − x ) − x 2 = x 2 − x + x − x . Thereforem ( x ) = − x 2 ( x 2 − x + x ) . m(x)=\frac{-x}{2(\sqrt{x^2-x}+x)}. m ( x ) = 2 ( x 2 − x + x ) − x . Limit as $x\to\infty$:Write ratio in terms of $\sqrt{x^2-x}/x = \sqrt{1-\frac{1}{x}} \to 1$. Thenx x 2 − x + x = 1 x 2 − x x + 1 → 1 1 + 1 = 1 2 . \frac{x}{\sqrt{x^2-x}+x} = \frac{1}{\frac{\sqrt{x^2-x}}{x}+1} \to \frac{1}{1+1}=\frac{1}{2}. x 2 − x + x x = x x 2 − x + 1 1 → 1 + 1 1 = 2 1 . Hencelim x → ∞ m ( x ) = − 1 2 ⋅ 1 2 = − 1 4 . \lim_{x\to\infty} m(x) = \frac{-1}{2}\cdot\frac{1}{2} = -\frac{1}{4}. lim x → ∞ m ( x ) = 2 − 1 ⋅ 2 1 = − 4 1 . Problem 6 Expression:p ( x ) = 4 + 3 e x 3 + e − x p(x)=\frac{4+3e^{x}}{3+e^{-x}} p ( x ) = 3 + e − x 4 + 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 → − ∞ p ( x ) = 0. \lim_{x\to-\infty} p(x)=0. lim x → − ∞ p ( x ) = 0. Problem 7 Expression:q ( x ) = 3 e 2 x + e − 3 x e 2 x − 4 e − 3 x q(x)=\frac{3e^{2x}+e^{-3x}}{e^{2x}-4e^{-3x}} q ( x ) = e 2 x − 4 e − 3 x 3 e 2 x + e − 3 x Factor $e^{2x}$:q ( x ) = e 2 x ( 3 + e − 5 x ) e 2 x ( 1 − 4 e − 5 x ) = 3 + e − 5 x 1 − 4 e − 5 x . q(x)=\frac{e^{2x}(3+e^{-5x})}{e^{2x}(1-4e^{-5x})} = \frac{3+e^{-5x}}{1-4e^{-5x}}. q ( x ) = e 2 x ( 1 − 4 e − 5 x ) e 2 x ( 3 + e − 5 x ) = 1 − 4 e − 5 x 3 + e − 5 x . Limit as $x\to\infty$:$e^{-5x}\to 0$; hence $q(x)\to 3/1 = 3$. Result:lim x → ∞ q ( x ) = 3. \lim_{x\to\infty} q(x)=3. lim x → ∞ q ( x ) = 3. Problem 8 Expression:r ( x ) = e x x + 3 , x → − 3 + . r(x)=\frac{e^{x}}{x+3},\qquad x\to-3^+. r ( x ) = x + 3 e x , x → − 3 + . As $x\to-3^+$: $x+3\to 0^+$, while $e^{x}\to e^{-3}>0$. Therefore the quotient blows up positively:lim x → − 3 + r ( x ) = + ∞ . \lim_{x\to-3^+} r(x)=+\infty. lim x → − 3 + r ( x ) = + ∞. Problem 9 Expression:s ( x ) = e x x + 3 , x → − 3 − . s(x)=\frac{e^{x}}{x+3},\qquad x\to-3^-. s ( x ) = x + 3 e x , x → − 3 − . As $x\to-3^-$: $x+3\to 0^-$, numerator $e^{x}>0$ stays finite. Therefore the quotient blows up negatively:lim x → − 3 − s ( x ) = − ∞ . \lim_{x\to-3^-} s(x)=-\infty. lim x → − 3 − s ( x ) = − ∞. Conclusion: vertical asymptote at $x=-3$ (with one-sided infinities as shown). Problem 10 Expression:t ( x ) = ln ( x + 4 3 x + 7 ) , x → ∞ . t(x)=\ln\left(\frac{x+4}{3x+7}\right),\quad x\to\infty. t ( x ) = ln ( 3 x + 7 x + 4 ) , x → ∞. Limit of the inside ratio:x + 4 3 x + 7 = 1 + 4 / x 3 + 7 / x → 1 3 . \frac{x+4}{3x+7} = \frac{1 + 4/x}{3 + 7/x} \to \frac{1}{3}. 3 x + 7 x + 4 = 3 + 7/ x 1 + 4/ x → 3 1 . Thereforelim x → ∞ t ( x ) = ln ( 1 3 ) = − ln 3. \lim_{x\to\infty} t(x) = \ln\left(\frac{1}{3}\right) = -\ln 3. lim x → ∞ t ( x ) = ln ( 3 1 ) = − ln 3. Problem 11 Expression:u ( x ) = ln ( x 2 − 9 ) − ln ( x 3 + 5 ) = ln ( x 2 − 9 x 3 + 5 ) . u(x)=\ln(x^2-9) - \ln(x^3+5) = \ln\left(\frac{x^2-9}{x^3+5}\right). u ( x ) = ln ( x 2 − 9 ) − ln ( x 3 + 5 ) = ln ( x 3 + 5 x 2 − 9 ) . 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 → ∞ [ ln ( x 2 − 9 ) − ln ( x 3 + 5 ) ] = − ∞ . \lim_{x\to\infty} [\ln(x^2-9) - \ln(x^3+5)] = -\infty. lim x → ∞ [ ln ( x 2 − 9 ) − ln ( x 3 + 5 )] = − ∞. Problem 12 Expression:v ( x ) = arctan ( x 2 − 9 x 2 + 3 x − 5 ) , x → ∞ . v(x)=\arctan\left(\frac{x^2-9}{x^2+3x-5}\right),\quad x\to\infty. v ( x ) = arctan ( x 2 + 3 x − 5 x 2 − 9 ) , x → ∞. Inside ratio:Degree both numerator and denominator is 2; leading terms give $\frac{x^2}{x^2}=1$, so the ratio $\to 1$ as $x\to\infty$. Thereforelim x → ∞ v ( x ) = arctan ( 1 ) = π 4 . \lim_{x\to\infty} v(x) = \arctan(1) = \frac{\pi}{4}. lim x → ∞ v ( x ) = arctan ( 1 ) = 4 π . Problem 13 Expression:w ( x ) = arctan ( x − 5 x − 3 ) , x → 3 + . w(x)=\arctan\left(\frac{x-5}{x-3}\right),\quad x\to 3^+. w ( x ) = arctan ( x − 3 x − 5 ) , x → 3 + . As $x\to 3^+$: numerator $(x-5)\to -2$, denominator $(x-3)\to 0^+$, so the fraction (\to -\infty). Thereforelim x → 3 + w ( x ) = arctan ( − ∞ ) = − π 2 . \lim_{x\to 3^+} w(x) = \arctan(-\infty) = -\frac{\pi}{2}. lim x → 3 + w ( x ) = arctan ( − ∞ ) = − 2 π . Problem 14 Expression:y ( x ) = arctan ( ln x ) , x → ∞ . y(x)=\arctan(\ln x),\quad x\to\infty. y ( x ) = arctan ( ln x ) , x → ∞. Behavior:$\ln x \to \infty$, so $\arctan(\ln x) \to \arctan(\infty) = \frac{\pi}{2}$. Result:lim x → ∞ arctan ( ln x ) = π 2 . \lim_{x\to\infty} \arctan(\ln x) = \frac{\pi}{2}. lim x → ∞ arctan ( ln x ) = 2 π . Summary of key ideas 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. Quick table of results (for quick reference) Problem 1: lim < e m > x → ∞ = 2 5 , lim < / e m > x → − ∞ = 2 5 , horizontal asymptote y = 2 5 . \lim<em>{x\to\infty}=\frac{2}{5},\quad \lim</em>{x\to-\infty}=\frac{2}{5},\quad \text{horizontal asymptote } y=\frac{2}{5}. lim < e m > x → ∞ = 5 2 , lim < / e m > x → − ∞ = 5 2 , horizontal asymptote y = 5 2 . Problem 2: lim x → ∞ = − ∞ ; no horizontal asymptote; oblique asymptote y = − 4 3 x − 14 9 . \lim_{x\to\infty}=-\infty;\quad \text{no horizontal asymptote; oblique asymptote } y=-\frac{4}{3}x-\frac{14}{9}. lim x → ∞ = − ∞ ; no horizontal asymptote; oblique asymptote y = − 3 4 x − 9 14 . Problem 3: lim x → − ∞ = 0. \lim_{x\to-\infty}=0. lim x → − ∞ = 0. Problem 4: lim x → − ∞ = − 3 5 . \lim_{x\to-\infty}= -\frac{3}{5}. lim x → − ∞ = − 5 3 . Problem 5: lim x → ∞ = − 1 4 . \lim_{x\to\infty}= -\frac{1}{4}. lim x → ∞ = − 4 1 . Problem 6: lim x → − ∞ = 0. \lim_{x\to-\infty}=0. lim x → − ∞ = 0. Problem 7: lim x → ∞ = 3. \lim_{x\to\infty}=3. lim x → ∞ = 3. Problem 8: lim x → − 3 + = + ∞ . \lim_{x\to-3^+}=+\infty. lim x → − 3 + = + ∞. Problem 9: lim x → − 3 − = − ∞ . \lim_{x\to-3^-}=-\infty. lim x → − 3 − = − ∞. Problem 10: lim x → ∞ = − ln 3. \lim_{x\to\infty}= -\ln 3. lim x → ∞ = − ln 3. Problem 11: lim x → ∞ = − ∞ . \lim_{x\to\infty}= -\infty. lim x → ∞ = − ∞. Problem 12: lim x → ∞ = π 4 . \lim_{x\to\infty}=\frac{\pi}{4}. lim x → ∞ = 4 π . Problem 13: lim x → 3 + = − π 2 . \lim_{x\to 3^+}= -\frac{\pi}{2}. lim x → 3 + = − 2 π . Problem 14: lim x → ∞ = π 2 . \lim_{x\to\infty}=\frac{\pi}{2}. lim x → ∞ = 2 π .