Calculus/L'Hôpital's rule/Solutions

L'Hôpital's rule Solutions

1. $\lim_{x \to 0}\frac{x+\tan x}{\sin x}$

$\lim_{x\to 0}\frac{1+\sec^2 x}{\cos x} = \mathbf{2}$

2. $\lim_{x \to \pi}\frac{x-\pi}{\sin x}$

$\lim_{x\to \pi}\frac{1}{\cos x} = \mathbf{-1}$

3. $\lim_{x \to 0}\frac{\sin 3x}{\sin 4x}$

$\lim_{x\to 0}\frac{3\cos(3x)}{4\cos(4x)} = \mathbf{\frac{3}{4}}$

4. $\lim_{x \to \infty}\frac{x^5}{e^{5x}}$

$\begin{align}\lim_{x\to \infty}\frac{5x^4}{5e^{5x}} &=\lim_{x\to \infty}\frac{5\cdot 4x^3}{5^2e^{5x}}\\ &=\lim_{x\to \infty}\frac{5\cdot 4\cdot 3x^2}{5^3e^{5x}}\\ &=\lim_{x\to \infty}\frac{5\cdot 4\cdot 3\cdot 2x}{5^4e^{5x}}\\ &=\lim_{x\to \infty}\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{5^5e^{5x}}\\ &=\mathbf{0}\end{align}$

5. $\lim_{x \to 0}\frac{\tan x - x}{\sin x - x}$

$\lim_{x\to 0}\frac{\sec^2 x-1}{\cos x-1}=\lim_{x\to 0}\frac{2\sec x \cos^{-2}x \sin x}{-\sin x}=\mathbf{-2}$