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How can I find a limit using equivalent functions and substitutions ( without applying L'Hospital) for the following problem?

There you can see the expression, I need to find the limit for, as x tends to 0

$\exp[(\cos(\sqrt x)-1)/x]$, square brackets for clarity.

amWhy
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user
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  • Hint: try using the power reduction formula for sine. – Jordan Green Jan 22 '19 at 18:42
  • Then it would be $$exp[(sinx)/x]$$ which is one? Could it be the solution of the problem be $$e^1$$? – user Jan 22 '19 at 18:54
  • Not quite. Note that $\frac{ \cos(\sqrt{x}) - 1}{x} = \frac{-2 (\sin(\sqrt{x}))^2}{x} = -2 (\frac{\sin(\sqrt{x})}{x})^2$. – Jordan Green Jan 22 '19 at 19:01
  • now I simplified: $$e^{-\frac{2\sin ^2\left(\sqrt{x}\right)}{x^2}}$$ Then It I would take the root of x in front of the expression it would fraction with -2 so the solution would be $$e^{-1}$$? – user Jan 22 '19 at 19:19
  • Sorry, I should have typed $\frac{-2(\sin(\sqrt{x})}{x} = -2 ( \frac{\sin(\sqrt{x})}{\sqrt{x}})^2$. What can you say about $\frac{\sin(\sqrt{x})}{\sqrt{x}}$ as $x$ tends to $0$ from the right? – Jordan Green Jan 22 '19 at 19:19
  • It tends to one? So then it's $$e^{-2}$$? – user Jan 22 '19 at 19:26
  • Yes, that’s it. – Jordan Green Jan 22 '19 at 19:30
  • Thank you for the answer. It seems as though I cannot either upvote you nor accept your comment as an answer... – user Jan 22 '19 at 20:02

1 Answers1

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You may set $x = t^2$ and use

  • $\lim_{y\to 0}\frac{1-\cos y}{y^2} = \frac{1}{2}$ (easy to verify with L'Hospital).

\begin{eqnarray*} e^{\frac{\cos\sqrt x -1}{x}} & = & e^{\frac{\cos t -1}{t^2}}\\ & \stackrel{t\to 0}{\longrightarrow} & e^{-\frac{1}{2}} =\frac{1}{\sqrt{e}}\\ \end{eqnarray*}

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