Tag: L’Hopital’s Rule

Compute the limit of the given function

by RoRi

January 11, 2016

Evaluate the limit.

$\lim_{x \to +\infty} \left( x^2 - \sqrt{x^4 -x^2 + 1} \right).$

First, we multiply and divide by the conjugate of the expression, then simplify and take the limit,

$\begin{align*} \lim_{x \to +\infty}\left( x^2 - \sqrt{x^4 - x^2 + 1} \right) &= \lim_{x \to +\infty} \left( \frac{\left(x^2 - \sqrt{x^4-x^2+1} \right)\left(x^2 + \sqrt{x^4-x^2+1} \right)}{x^2 + \sqrt{x^4-x^2+1}} \right)\\[9pt] &= \lim_{x \to +\infty} \frac{x^4 - x^4 + x^2 - 1}{x^2 + \sqrt{x^4 - x^2 + 1}} \\[9pt] &= \lim_{x \to +\infty} \frac{x^2-1}{x^2 + x^2 \sqrt{1 - \frac{1}{x^2} + \frac{1}{x^4}}} \\[9pt] &= \lim_{x \to +\infty} \frac{1 - \frac{1}{x^2}}{1 + \sqrt{1-\frac{1}{x^2} + \frac{1}{x^4}}} \\[9pt] &= \frac{1}{2}. \end{align*}$

Compute the limit of the given function

by RoRi

January 11, 2016

Evaluate the limit.

$\lim_{x \to +\infty} x^{\frac{1}{4}} \sin \left( \frac{1}{\sqrt{x}} \right).$

First, we make the substitution $t = \frac{1}{\sqrt{x}}$ . Then $t \to 0^+$ as $x \to +\infty$ so we have

$\begin{align*} \lim_{x \to +\infty} x^{\frac{1}{4}} \sin \left( \frac{1}{\sqrt{x}} \right) &= \lim_{t \to 0^+} \frac{\sin t}{\sqrt{t}} \\[9pt] &= \lim_{t \to 0^+} \frac{\cos t}{\frac{1}{2\sqrt{t}}} &(\text{L'Hopital's})\\[9pt] &= \lim_{t \to 0^+} 2 \sqrt{t} \cos t \\[9pt] &= 0. \end{align*}$

Compute the limit of the given function

by RoRi

December 25, 2015

Evaluate the limit.

$\lim_{x \to 0^+} \frac{1}{\sqrt{x}} \left( \frac{1}{\sin x} - \frac{1}{x} \right).$

First,

$\frac{1}{\sqrt{x}} \left( \frac{1}{\sin x} - \frac{1}{x} \right) = \frac{1}{\sqrt{x}} \cdot \frac{x - \sin x}{x \sin x} = \frac{x - \sin x}{x^{\frac{3}{2}} \sin x}.$

Then, we multiply inside the limit by $\frac{\sin x}{x}$ since $\lim_{x \to 0} \frac{\sin x}{x} = 1$ ,

$\begin{align*} \lim_{x \to 0^+} \frac{1}{\sqrt{x}} \left(\frac{1}{\sin x} - \frac{1}{x} \right) &= \lim_{x \to 0^+} \frac{x - \sin x}{x^{\frac{3}{2}} \sin x} \\[9pt] &= \lim_{x \to 0^+} \left( \frac{x - \sin x}{x^{\frac{3}{2}} \sin x} \cdot \frac{\sin x}{x} \right) \\[9pt] &= \lim_{x \to 0^+} \frac{x - \sin x}{x^{\frac{5}{2}}} \\[9pt] &= \lim_{x \to 0^+} \frac{1 - \cos x}{\frac{5}{2} x^{\frac{3}{2}}} &(\text{L'Hopital's})\\[9pt] &= \lim_{x \to 0^+} \frac{4 \sin x}{15 \sqrt{x}} &(\text{L'Hopital's})\\[9pt] &= \lim_{x \to 0^+} \frac{8 \sqrt{x} \cos x}{15} &(\text{L'Hopital's})\\[9pt] &= 0. \end{align*}$

Compute the limit of the given function

by RoRi

December 25, 2015

Evaluate the limit.

$\lim_{x \to + \infty} \frac{\cosh (x+1)}{e^x}.$

We recall the definition of the hyperbolic cosine in terms of the exponential,

$\cosh (x+1) = \frac{e^{x+1} + e^{-(x+1)}}{2}.$

Using this we compute,

$\begin{align*} \lim_{x \to +\infty} \frac{\cosh(x+1)}{e^x} &= \lim_{x \to +\infty} \frac{e^{x+1} + e^{-(x+1)}}{2e^x} \\[9pt] &= \lim_{x \to + \infty} \left( \frac{e}{2} + \frac{1}{2e^{2x+1}} \right) \\[9pt] &= \frac{e}{2}. \end{align*}$

Compute the limit of the given function

by RoRi

December 25, 2015

Evaluate the limit.

$\lim_{x \to \frac{1}{2}^-} \frac{\log(1-2x)}{\tan (\pi x)}.$

Both the numerator and denominator go to 0 as $x \to \frac{1}{2}^-$ so we apply L’Hopital’s rule,

$\begin{align*} \lim_{x \to \frac{1}{2}^-} \frac{\log (1 - 2x)}{\tan (\pi x)} &= \lim_{x \to \frac{1}{2}^-} \frac{ \frac{-2}{1-2x} }{ \frac{\pi}{\cos^2 (\pi x)} } \\[9pt] &= \lim_{x \to \frac{1}{2}^-} \frac{-2 \cos^2 (\pi x)}{\pi (1-2x)} \\[9pt] &= \lim_{x \to \frac{1}{2}^-} \frac{(-4 \cos (\pi x)) (\pi) (- \sin (\pi x))}{-2 \pi} &(\text{L'Hopital's})\\[9pt] &= 0. \end{align*}$

Compute the limit of the given function

by RoRi

December 25, 2015

Evaluate the limit.

$\lim_{x \to \pi} \frac{\log |\sin x|}{\log | \sin (2x) |}.$

We use the trig identity $\sin (2x) = 2 \sin x \cos x$ and then use L’Hopital’s rule to evaluate,

$\begin{align*} \lim_{x \to \pi} \frac{\log |\sin x|}{\log |\sin (2x)|} &= \lim_{x \to \pi} \frac{\log |\sin x|}{\log |2 \sin x \cos x|} \\[9pt] &= \lim_{x \to \pi} \frac{\log |\sin x|}{\log 2 + \log |\sin x| + \log |\cos x|} \\[9pt] &= \lim_{x \to \pi} \frac{ \frac{\cos x}{\sin x} }{ \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}} &(\text{L'Hopital's}) \\[9pt] &= \lim_{x \to \pi} \frac{\cos x}{\cos x - \frac{\sin x}{\cos^2 x}} \\[9pt] &= 1. \end{align*}$

Compute the limit of the given function

by RoRi

December 25, 2015

Evaluate the limit.

$\lim_{x \to +\infty} x^4 \left( \cos \frac{1}{x} - 1 + \frac{1}{2x^2} \right).$

Let $t = \frac{1}{x}$ , then $t \to 0$ as $x \to +\infty$ and we have

$\begin{align*} \lim_{x \to +\infty} x^4 \left( \cos \frac{1}{x} - 1 + \frac{1}{2x^2} \right) &= \lim_{t \to 0^+} \left( \frac{\cos t}{t^4} - \frac{1}{t^4} + \frac{1}{2t^2} \right)\\[9pt] &= \lim_{t \to 0^+} \frac{ 2 \cos t - 2 + t^2}{2t^4} \\[9pt] &= \lim_{t \to 0^+} \frac{ 2t - 2 \sin t}{8t^3} &(\text{L'Hopital's})\\[9pt] &= \lim_{t \to 0^+} \frac{ 2 - 2 \cos t}{24t^2} &(\text{L'Hopital's})\\[9pt] &= \lim_{t \to 0^+} \frac{ \sin t}{24t} &(\text{L'Hopital's})\\[9pt] &= \lim_{t \to 0^+} \frac{\cos t}{24} &(\text{L'Hopital's})\\[9pt] &= \frac{1}{24}. \end{align*}$

Compute the limit of the given function

by RoRi

December 25, 2015

Evaluate the limit.

$\lim_{x \to +\infty} \frac{\log (a + be^x)}{\sqrt{a+bx^2}}.$

First, we pull an $e^x$ out of $(a+be^x)$ and an $x^2$ out of $a+bx^2$ to write

$\frac{\log(a+be^x)}{\sqrt{a+bx^2}} = \frac{\log((e^x)(ae^{-x} + b)}{x \sqrt{ax^{-2} + b}} = \frac{x + \log(ae^{-x} + b)}{x \sqrt{ax^{-2} + b}}.$

Then we have

$\begin{align*} \lim_{x \to +\infty} \frac{\log (a + be^x)}{\sqrt{a+bx^2}} &= \lim_{x \to +\infty} \frac{x + \log(ae^{-x} +b)}{x\sqrt{ax^{-2} + b}} \\[9pt] &= \lim_{x \to +\infty} \left( \frac{1}{\sqrt{ax^{-2} + b}} + \frac{\log(ae^{-x} + b)}{x \sqrt{ax^{-2} + b}} \right)\\[9pt] &= \frac{1}{\sqrt{b}} + 0 &(\text{Theorem 7.11})\\ &= \frac{1}{\sqrt{b}}. \end{align*}$

Compute the limit of the given function

by RoRi

December 25, 2015

Evaluate the limit.

$\lim_{x \to \frac{\pi}{2}} \frac{\tan (3x)}{\tan x}.$

We write $\tan x = \frac{\sin x}{\cos x}$ and apply L’Hopital’s rule to solve

$\begin{align*} \lim_{x \to \frac{\pi}{2}} \frac{\tan (3x)}{\tan x} &= \lim_{x \to \frac{\pi}{2}} \left( \frac{\sin (3x)}{\cos (3x)} \cdot \frac{\cos x}{\sin x} \right) \\[9pt] &= \lim_{x \to \frac{\pi}{2}} \frac{ 3 \cos x \cos (3x) - \sin x \sin (3x)}{\cos x \cos(3x) - 3 \sin x \sin (3x)} \\[9pt] &= \frac{1}{3}. \end{align*}$

Compute the limit of the given function

by RoRi

December 25, 2015

Evaluate the limit.

$\lim_{x \to +\infty} \frac{\sin \frac{1}{x}}{\arctan \frac{1}{x}}.$

Let $t = \frac{1}{x}$ , then $t \to 0$ as $x \to +\infty$ . Making this substitution and using L’Hopital’s rule we have

$\begin{align*} \lim_{x \to +\infty} \frac{\sin \frac{1}{x}}{\arctan \frac{1}{x}} &= \lim_{t \to 0} \frac{\sin t}{\arctan t} \\[9pt] &= \lim_{t \to 0} \frac{\cos t}{\frac{1}{1+t^2}} \\[9pt] &= 1. \end{align*}$