L'Hôspital Rule

Suppose $f (x)$ and $g (x)$ are differentiable and $g^{'} (x) \neq 0$ on an open interval $ℐ$ that contains $ω$ except possibly at $ω$ . Suppose

Then

lim_{x \to ω} \frac{f (x)}{g (x)} = \frac{f' (x)}{g' (x)}

provided the limit on the right exists.

$0 / 0$

\begin{matrix} lim_{x \to 0} \frac{1 - cos x}{x} \\ lim_{x \to 1} \frac{sin (π x)}{ln x} \\ lim_{x \to \infty} \frac{ℯ^{\frac{1}{x}} - 1}{\frac{1}{x}} \\ lim_{x \to 0} \frac{sin x - x}{x} \end{matrix}

\begin{matrix} lim_{x \to \infty} \frac{3 x + 5}{2 x - 1} \\ lim_{x \to 0^{+}} \frac{ln (x)}{1 / tan (x)} \\ lim_{x \to \infty} \frac{ln (x)}{x} \\ lim_{x \to \infty} \frac{ℯ^{x}}{x} \end{matrix}

\begin{matrix} lim_{x \to 1} \frac{x^{2} + 5}{3 x - 4} \\ lim_{x \to \infty} \frac{x - sin x}{x} \end{matrix}

\begin{matrix} lim_{x \to 0^{+}} (x ln x) \end{matrix}

\begin{matrix} lim_{x \to 0^{-}} (\frac{1}{x} - \frac{1}{sin x}) \\ lim_{x \to 0^{+}} (\frac{1}{x^{2}} - \frac{1}{tan x}) \end{matrix}

\begin{matrix} lim_{x \to \infty} x^{\frac{1}{x}} \end{matrix}

\begin{matrix} lim_{x \to 0^{+}} x^{x} \\ lim_{x \to 0^{+}} x^{sin x} \end{matrix}

\begin{matrix} lim_{x \to 0^{+}} {(1 + x)}^{\frac{1}{x}} \end{matrix}