# Calculus/L'Hôpital's rule/Solutions

## L'Hôpital's rule Solutions

1. ${\displaystyle \lim _{x\to 0}{\frac {x+\tan x}{\sin x}}}$
${\displaystyle \lim _{x\to 0}{\frac {1+\sec ^{2}x}{\cos x}}=\mathbf {2} }$
${\displaystyle \lim _{x\to 0}{\frac {1+\sec ^{2}x}{\cos x}}=\mathbf {2} }$
2. ${\displaystyle \lim _{x\to \pi }{\frac {x-\pi }{\sin x}}}$
${\displaystyle \lim _{x\to \pi }{\frac {1}{\cos x}}=\mathbf {-1} }$
${\displaystyle \lim _{x\to \pi }{\frac {1}{\cos x}}=\mathbf {-1} }$
3. ${\displaystyle \lim _{x\to 0}{\frac {\sin 3x}{\sin 4x}}}$
${\displaystyle \lim _{x\to 0}{\frac {3\cos(3x)}{4\cos(4x)}}=\mathbf {\frac {3}{4}} }$
${\displaystyle \lim _{x\to 0}{\frac {3\cos(3x)}{4\cos(4x)}}=\mathbf {\frac {3}{4}} }$
4. ${\displaystyle \lim _{x\to \infty }{\frac {x^{5}}{e^{5x}}}}$
{\displaystyle {\begin{aligned}\lim _{x\to \infty }{\frac {5x^{4}}{5e^{5x}}}&=\lim _{x\to \infty }{\frac {5\cdot 4x^{3}}{5^{2}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3x^{2}}{5^{3}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3\cdot 2x}{5^{4}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3\cdot 2\cdot 1}{5^{5}e^{5x}}}\\&=\mathbf {0} \end{aligned}}}
{\displaystyle {\begin{aligned}\lim _{x\to \infty }{\frac {5x^{4}}{5e^{5x}}}&=\lim _{x\to \infty }{\frac {5\cdot 4x^{3}}{5^{2}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3x^{2}}{5^{3}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3\cdot 2x}{5^{4}e^{5x}}}\\&=\lim _{x\to \infty }{\frac {5\cdot 4\cdot 3\cdot 2\cdot 1}{5^{5}e^{5x}}}\\&=\mathbf {0} \end{aligned}}}
5. ${\displaystyle \lim _{x\to 0}{\frac {\tan x-x}{\sin x-x}}}$
${\displaystyle \lim _{x\to 0}{\frac {\sec ^{2}x-1}{\cos x-1}}=\lim _{x\to 0}{\frac {2\sec x\cos ^{-2}x\sin x}{-\sin x}}=\mathbf {-2} }$
${\displaystyle \lim _{x\to 0}{\frac {\sec ^{2}x-1}{\cos x-1}}=\lim _{x\to 0}{\frac {2\sec x\cos ^{-2}x\sin x}{-\sin x}}=\mathbf {-2} }$