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How to solve this limit $$\lim_{x\to0^+}\cos(\sqrt{x})^{1/x}$$ without L'Hospital's rule.

Bernard
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Robin Pastrňák
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  • By writing it as $\exp\left(\frac{\ln(\cos(\sqrt{x}))}{x}\right)$ you can reduce it to computing the limit of $\frac{\ln(\cos(\sqrt{x}))}{x}=\frac{\ln(1+(\cos(\sqrt{x})-1))}{\cos(\sqrt{x})-1}\frac{\cos(\sqrt{x})-1}{x}$. Use for these that $\lim_{x\to0}\frac{\ln(1+x)}{x}=1$ and that $\lim_{x\to0}\frac{1-\cos(x)}{x^2/2}=1$. – conditionalMethod Nov 20 '19 at 15:26

3 Answers3

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Using the series expansion of $~\cos~$ (Taylor series at $x=0$):

$\displaystyle\cos(\sqrt{x})^{1/x}\approx\left(1-\frac{x}{2}\right)^{1/x}\approx e^{-1/2}$

user90369
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HINT

Use that

$$\cos(\sqrt{x})^{1/x}=\left[(1+(\cos(\sqrt{x})-1))^{\frac{1}{\cos(\sqrt{x})-1}}\right]^\frac{\cos(\sqrt{x})-1}{x}$$

and the relevant standard limits.

user
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Let $\sqrt x=2y$

so that the limit becomes $$\left(\lim_{y\to0^+}(1-2\sin^2y)^{-1/2\sin^2y}\right)^{-2\sin^2y/4y^2}$$

$$=e^{-1/2}$$

lab bhattacharjee
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