L'Hopital's Rule

Extended Mean Value Theorem

The Extended Mean Value Theorem says that if f and g are differentiable on (a,b) there exists a point c $\in$ (a,b) s.t. $\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$ Recall the regular mean value theorem, with $g(x) = x$: $\frac{f'(c)}{x'} = \frac{f(b) - f(a)}{b - a}$ $x' = 1$, so $f'(c) = \frac{f(b) - f(a)}{b - a}$

L'Hôpital's Rule

$\text{Let's say you have} \lim_{x \to c}{\frac{f(x)}{g(x)}\text{ , and that evaluates to} \frac{0}{0}, \frac{\infty}{\infty}, \frac{-\infty}{\infty}, \frac{\infty}{-\infty}, \text{ or } \frac{-\infty}{-\infty}}$
$\text{That is equivalent to }\lim_{x \to c}{\frac{f'(x)}{g'(x)}}$

Example (0/0)

$$\lim_{x \to 0}{\frac{e^{2x} - 1}{x}} = \frac{0}{0}$$ $$\lim_{x \to 0}{\frac{2e^{2x}}{1}} = 2$$

Example (∞/∞)

$$\lim_{x\to \infty}{\frac{\ln{x}}{x}} = \frac{\infty}{\infty}$$ $$\lim_{x\to \infty}{\frac{1}{x}} = 0$$

Example (∞ - ∞)

$$\lim_{x \to 1}{\frac{1}{\ln{x}} - \frac{1}{x-1}}$$ $$=\lim_{x \to 1}{\frac{x - 1 - \ln{x}}{(x-1)\ln{x}}}$$ $$=\lim_{x \to 1}{\frac{1 - \frac{1}{x}}{\ln{x} + \frac{x-1}{x}}}$$ $$(\frac{x-1}{x} = 1 - \frac{1}{x})$$ $$\lim_{x \to 1}{\frac{\frac{1}{x^2}}{\frac{1}{x} + \frac{1}{x^2}}}$$ $$=\lim_{x \to 1}{\frac{1}{1 + 1}} = \frac{1}{2}$$

Example (0^0)

$$y = \lim_{x \to 0}{\sin^x{x}}$$ $$\ln{y} = \ln{\lim_{x \to 0}{sin^x{x}}}$$ $$\ln{y} = \lim_{x \to 0}{\ln{sin^x{x}}}$$ $$\ln{y} = \lim_{x \to 0}{x\ln{sin{x}}}$$ $$\ln{y} = \lim_{x \to 0}{\frac{\ln{sin{x}}}{\frac{1}{x}}}$$ $$\text{L'H: }\ln{y} = \lim_{x \to 0}{-\cot{(x)}x^2}$$ $$\ln{y} = \lim_{x \to 0}{-\frac{x^2}{\tan{x}}}$$ $$\ln{y} = \lim_{x \to 0}{-\frac{2x}{\sec^2{x}}}$$ $$\ln{y} = \lim_{x \to 0}{-2x\cos^2{x}}$$ $$\ln{y} = 0$$ $$y = 1$$

Example (∞^0)

$$y = \lim_{x \to \infty}{x^{\frac{1}{x}}}$$ $$\text{Now, do the same thing as } 0^0 \text{, take the natural log}$$ $$\ln{y} = \lim_{x \to \infty}{\ln{(x^{\frac{1}{x}}})}$$ $$\ln{y} = \lim_{x \to \infty}{\frac{\ln{(x)}}{x}}$$ $$\ln{y} = \lim{\frac{\frac{1}{x}}{1}}$$ $$\ln{y} = 0$$ $$y = 1$$