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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 : $\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e$

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

1 Answer 1

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$$\begin{align} \lim_{x \to 0} (\cos x)^{\cot x} & =\lim_{x \to 0} ((1+(\cos x-1))^{\frac{1}{\cos x -1}})^{(\cos x -1)\cot x}\\\\ & =\lim_{x \to 0} e^{(\cos x -1)\cot x}\\\\ & =e^{{\lim_{x \to 0} {(\cos x -1)\cot x}}}\\\\& =e^0\\\\ &=1 \end{align}$$

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    $\begingroup$ thank you kind sire $\endgroup$ Jul 3, 2023 at 16:07
  • $\begingroup$ @SamarthSaluja you're welcome. $\endgroup$
    – bb_823
    Jul 3, 2023 at 16:07
  • $\begingroup$ 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 $\lim f(x)^{g(x)} = \lim e^{g(x)}$. $\endgroup$
    – jjagmath
    Jul 3, 2023 at 16:17
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    $\begingroup$ 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, $\;\lim [f(x)^{g(x)}]=[\lim f(x)]^{\lim g(x)}.$ $\endgroup$
    – Angelo
    Jul 3, 2023 at 16:22
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    $\begingroup$ @bb_823, good answer (+1). $\endgroup$
    – Angelo
    Jul 3, 2023 at 16:27

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