L’Hôpital’s Rule is a calculus technique for evaluating limits that produce indeterminate forms. Instead of trying to simplify the original expression directly, you take the derivative of the numerator and denominator separately and then evaluate the new limit. It is especially helpful for limits involving logarithms, exponentials, trigonometric functions, and rational expressions.
Notes
Questions
\(\textbf{1)}\) \( \displaystyle \lim_{x\to1} \displaystyle \frac{\ln{x}}{x-1} \)
The answer is \(1\)
\(\,\,\,\,\,\,\text{Verify }\displaystyle \frac {\ln{(1)}}{1-1}= \frac{0}{0}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to c} \displaystyle \frac{f(x)}{g(x)} = \lim_{x\to c} \displaystyle \frac{f'(x)}{g'(x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to 1} \displaystyle \frac{\ln{x}}{x-1} = \lim_{x\to 1} \displaystyle \frac{1/x}{1}\)
\(\,\,\,\,\,\, \lim_{x\to 1} \frac{1/x}{1}=\frac{1/(1)}{1}=1\)
\(\,\,\,\,\,\,\,\text{The answer is } 1\)
\(\textbf{2)}\) \( \displaystyle \lim_{x\to\infty} \displaystyle \frac{\ln{x}}{x} \)
The answer is \(0\)
\(\,\,\,\,\,\text{Verify this has the form }\displaystyle \frac{\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{\ln{x}}{x}=\lim_{x\to\infty}\frac{1/x}{1}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{1}{x}=0\)
\(\,\,\,\,\,\text{The answer is }0\)
\(\textbf{3)}\) \( \displaystyle \lim_{x\to\infty} xe^{-x} \)
The answer is \(0\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}xe^{-x}=\lim_{x\to\infty}\frac{x}{e^x}\)
\(\,\,\,\,\,\text{Verify this has the form }\displaystyle \frac{\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{x}{e^x}=\lim_{x\to\infty}\frac{1}{e^x}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{1}{e^x}=0\)
\(\,\,\,\,\,\text{The answer is }0\)
\(\textbf{4)}\) \( \displaystyle \lim_{x\to0} (\sin x)^x \)
The answer is \(1\)
\(\,\,\,\,\,\text{Let }L=\displaystyle \lim_{x\to0^+}(\sin x)^x\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to0^+}x\ln(\sin x)\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to0^+}\frac{\ln(\sin x)}{1/x}\)
\(\,\,\,\,\,\text{This has the form }\displaystyle \frac{-\infty}{\infty}\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to0^+}\frac{\cot x}{-1/x^2}\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to0^+}-x^2\cot x\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to0^+}\frac{-x^2\cos x}{\sin x}=0\)
\(\,\,\,\,\,L=e^0=1\)
\(\,\,\,\,\,\text{The answer is }1\)
\(\textbf{5)}\) \( \displaystyle \lim_{x\to\infty} \displaystyle \frac{e^x}{x^3} \)
The answer is \(\infty\)
\(\,\,\,\,\,\text{Verify this has the form }\displaystyle \frac{\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{e^x}{x^3}=\lim_{x\to\infty}\frac{e^x}{3x^2}\)
\(\,\,\,\,\,\displaystyle =\lim_{x\to\infty}\frac{e^x}{6x}\)
\(\,\,\,\,\,\displaystyle =\lim_{x\to\infty}\frac{e^x}{6}=\infty\)
\(\,\,\,\,\,\text{The answer is }\infty\)
\(\textbf{6)}\) \( \displaystyle \lim_{x\to\infty} \displaystyle \frac{5x^2+2x-3}{4x^2-10x+7} \)
The answer is \(\displaystyle\frac{5}{4}\)
\(\,\,\,\,\,\text{Verify this has the form }\displaystyle \frac{\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{5x^2+2x-3}{4x^2-10x+7}=\lim_{x\to\infty}\frac{10x+2}{8x-10}\)
\(\,\,\,\,\,\displaystyle =\lim_{x\to\infty}\frac{10}{8}\)
\(\,\,\,\,\,\displaystyle \frac{10}{8}=\frac{5}{4}\)
\(\,\,\,\,\,\text{The answer is }\frac{5}{4}\)
\(\textbf{7)}\) \( \displaystyle \lim_{x\to\infty} (1+ \frac{1}{x})^x \)
The answer is \(e\)
\(\,\,\,\,\,\text{Let }L=\displaystyle \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to\infty}x\ln\left(1+\frac{1}{x}\right)\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{1/x}\)
\(\,\,\,\,\,\text{This has the form }\displaystyle \frac{0}{0}\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to\infty}\frac{\frac{-1/x^2}{1+1/x}}{-1/x^2}\)
\(\,\,\,\,\,\ln L=\displaystyle \lim_{x\to\infty}\frac{1}{1+1/x}=1\)
\(\,\,\,\,\,L=e^1=e\)
\(\,\,\,\,\,\text{The answer is }e\)
\(\textbf{8)}\) \( \displaystyle \lim _{x\to 0}\frac{\sin x}{x} \)
The answer is \(1\)
\(\,\,\,\,\,\text{Verify }\displaystyle \frac{\sin(0)}{0}=\frac{0}{0}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\cos x}{1}\)
\(\,\,\,\,\,\displaystyle \frac{\cos(0)}{1}=1\)
\(\,\,\,\,\,\text{The answer is }1\)
\(\textbf{9)}\) \( \displaystyle \lim_{x\to0}\frac{e^x-1}{x} \)
The answer is \(1\)
\(\,\,\,\,\,\text{Verify }\displaystyle \frac{e^0-1}{0}=\frac{0}{0}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{e^x}{1}\)
\(\,\,\,\,\,\displaystyle \frac{e^0}{1}=1\)
\(\,\,\,\,\,\text{The answer is }1\)
\(\textbf{10)}\) \( \displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2} \)
The answer is \(\frac{1}{2}\)
\(\,\,\,\,\,\text{Verify }\displaystyle \frac{1-\cos(0)}{0^2}=\frac{0}{0}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{x\to0}\frac{\sin x}{2x}\)
\(\,\,\,\,\,\text{This is still }\displaystyle \frac{0}{0}\text{, so use L’Hôpital’s Rule again.}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin x}{2x}=\lim_{x\to0}\frac{\cos x}{2}\)
\(\,\,\,\,\,\displaystyle \frac{\cos(0)}{2}=\frac{1}{2}\)
\(\,\,\,\,\,\text{The answer is }\frac{1}{2}\)
\(\textbf{11)}\) \( \displaystyle \lim_{x\to\infty}\frac{x^2}{e^x} \)
The answer is \(0\)
\(\,\,\,\,\,\text{Verify this has the form }\displaystyle \frac{\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{x^2}{e^x}=\lim_{x\to\infty}\frac{2x}{e^x}\)
\(\,\,\,\,\,\text{This is still }\displaystyle \frac{\infty}{\infty}\text{, so use L’Hôpital’s Rule again.}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{2x}{e^x}=\lim_{x\to\infty}\frac{2}{e^x}=0\)
\(\,\,\,\,\,\text{The answer is }0\)
\(\textbf{12)}\) \( \displaystyle \lim_{x\to0}\frac{\ln(1+x)}{x} \)
The answer is \(1\)
\(\,\,\,\,\,\text{Verify }\displaystyle \frac{\ln(1+0)}{0}=\frac{0}{0}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\ln(1+x)}{x}=\lim_{x\to0}\frac{1/(1+x)}{1}\)
\(\,\,\,\,\,\displaystyle \frac{1/(1+0)}{1}=1\)
\(\,\,\,\,\,\text{The answer is }1\)
\(\textbf{13)}\) \( \displaystyle \lim_{x\to\infty}\frac{3x^3+2}{5x^3-x} \)
The answer is \(\frac{3}{5}\)
\(\,\,\,\,\,\text{Verify this has the form }\displaystyle \frac{\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{3x^3+2}{5x^3-x}=\lim_{x\to\infty}\frac{9x^2}{15x^2-1}\)
\(\,\,\,\,\,\displaystyle =\lim_{x\to\infty}\frac{18x}{30x}\)
\(\,\,\,\,\,\displaystyle =\lim_{x\to\infty}\frac{18}{30}=\frac{3}{5}\)
\(\,\,\,\,\,\text{The answer is }\frac{3}{5}\)
\(\textbf{14)}\) \( \displaystyle \lim_{x\to0}\frac{\tan x-x}{x^3} \)
The answer is \(\frac{1}{3}\)
\(\,\,\,\,\,\text{Verify }\displaystyle \frac{\tan(0)-0}{0^3}=\frac{0}{0}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{\sec^2 x-1}{3x^2}\)
\(\,\,\,\,\,\text{This is still }\displaystyle \frac{0}{0}\text{, so use L’Hôpital’s Rule again.}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sec^2 x-1}{3x^2}=\lim_{x\to0}\frac{2\sec^2 x\tan x}{6x}\)
\(\,\,\,\,\,\text{This is still }\displaystyle \frac{0}{0}\text{, so use L’Hôpital’s Rule again.}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{2\sec^2 x\tan x}{6x}=\lim_{x\to0}\frac{4\sec^2 x\tan^2 x+2\sec^4 x}{6}\)
\(\,\,\,\,\,\displaystyle \frac{4(1)(0)^2+2(1)^2}{6}=\frac{1}{3}\)
\(\,\,\,\,\,\text{The answer is }\frac{1}{3}\)
\(\textbf{15)}\) \( \displaystyle \lim_{x\to0}\frac{e^x-1-x}{x^2} \)
The answer is \(\frac{1}{2}\)
\(\,\,\,\,\,\text{Verify }\displaystyle \frac{e^0-1-0}{0^2}=\frac{0}{0}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{e^x-1}{2x}\)
\(\,\,\,\,\,\text{This is still }\displaystyle \frac{0}{0}\text{, so use L’Hôpital’s Rule again.}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{e^x-1}{2x}=\lim_{x\to0}\frac{e^x}{2}\)
\(\,\,\,\,\,\displaystyle \frac{e^0}{2}=\frac{1}{2}\)
\(\,\,\,\,\,\text{The answer is }\frac{1}{2}\)
\(\textbf{16)}\) \( \displaystyle \lim_{x\to0}\frac{\sin(3x)}{\sin(5x)} \)
The answer is \(\frac{3}{5}\)
\(\,\,\,\,\,\text{Verify }\displaystyle \frac{\sin(0)}{\sin(0)}=\frac{0}{0}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin(3x)}{\sin(5x)}=\lim_{x\to0}\frac{3\cos(3x)}{5\cos(5x)}\)
\(\,\,\,\,\,\displaystyle \frac{3\cos(0)}{5\cos(0)}=\frac{3}{5}\)
\(\,\,\,\,\,\text{The answer is }\frac{3}{5}\)
\(\textbf{17)}\) \( \displaystyle \lim_{x\to\infty}\frac{\ln(x^2)}{x} \)
The answer is \(0\)
\(\,\,\,\,\,\text{Verify this has the form }\displaystyle \frac{\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{\ln(x^2)}{x}=\lim_{x\to\infty}\frac{2/x}{1}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{2}{x}=0\)
\(\,\,\,\,\,\text{The answer is }0\)
\(\textbf{18)}\) \( \displaystyle \lim_{x\to0^+}x\ln x \)
The answer is \(0\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0^+}x\ln x=\lim_{x\to0^+}\frac{\ln x}{1/x}\)
\(\,\,\,\,\,\text{This has the form }\displaystyle \frac{-\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0^+}\frac{\ln x}{1/x}=\lim_{x\to0^+}\frac{1/x}{-1/x^2}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0^+}(-x)=0\)
\(\,\,\,\,\,\text{The answer is }0\)
\(\textbf{19)}\) \( \displaystyle \lim_{x\to0}\frac{e^{2x}-1}{\sin x} \)
The answer is \(2\)
\(\,\,\,\,\,\text{Verify }\displaystyle \frac{e^0-1}{\sin(0)}=\frac{0}{0}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{e^{2x}-1}{\sin x}=\lim_{x\to0}\frac{2e^{2x}}{\cos x}\)
\(\,\,\,\,\,\displaystyle \frac{2e^0}{\cos(0)}=2\)
\(\,\,\,\,\,\text{The answer is }2\)
\(\textbf{20)}\) \( \displaystyle \lim_{x\to\infty}\frac{x}{e^x} \)
The answer is \(0\)
\(\,\,\,\,\,\text{Verify this has the form }\displaystyle \frac{\infty}{\infty}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{x}{e^x}=\lim_{x\to\infty}\frac{1}{e^x}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to\infty}\frac{1}{e^x}=0\)
\(\,\,\,\,\,\text{The answer is }0\)
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