# $\lim_{x \to 0} (\cos x)^{\cot x}$

The following question is from cengage calculus . Illustration 2.95 but the explanation isn't clear to me

$$\lim_{x \to 0} (\cos x)^{\cot x}$$

It is to be solved by using the identity : $$\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e$$

• $\lim_{x \to 0} (1+x)^{\frac{1}{x}}$ is not an identity. You missed "$= e$" Jul 3, 2023 at 16:10
• my bad. was in a hurry Jul 3, 2023 at 16:13

\begin{align} \lim_{x \to 0} (\cos x)^{\cot x} & =\lim_{x \to 0} ((1+(\cos x-1))^{\frac{1}{\cos x -1}})^{(\cos x -1)\cot x}\\\\ & =\lim_{x \to 0} e^{(\cos x -1)\cot x}\\\\ & =e^{{\lim_{x \to 0} {(\cos x -1)\cot x}}}\\\\& =e^0\\\\ &=1 \end{align}
• You can't apply a limit to just part of an expression. Specifically in this case, from $\lim f(x) = e$ you can't conclude $\lim f(x)^{g(x)} = \lim e^{g(x)}$. Jul 3, 2023 at 16:17
• The function $F(u,v)=u^v$ is continuous on $\Bbb R^+\!\times\Bbb R$, so if $u\to e$ and $v\to0$, then $u^v\to e^0=1$. Consequently, $\;\lim [f(x)^{g(x)}]=[\lim f(x)]^{\lim g(x)}.$ Jul 3, 2023 at 16:22