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Question Number 129211 by bramlexs22 last updated on 13/Jan/21

$$\underset{x\to 0}{\lim }\frac{1}{\cos {\left(\sqrt{\mathrm{x}}\right)}^{\frac{1}{\mathrm{x}}}}=?$$

Answered by liberty last updated on 13/Jan/21

$$\mathcal{L}=\underset{x\to 0}{\lim }\frac{1}{\cos {\left(\sqrt{\mathrm{x}}\right)}^{\frac{1}{\mathrm{x}}}}$$$$\ln \mathcal{L}=\underset{x\to 0}{\lim }\ln \left(\frac{1}{\cos {\left(\sqrt{\mathrm{x}}\right)}^{\frac{1}{\mathrm{x}}}}\right)$$$$\ln \mathcal{L}=\underset{x\to 0}{\lim }-\frac{\ln \cos \left(\sqrt{\mathrm{x}}\right)}{\mathrm{x}};\left[{\mathrm{L}}^{\prime }\mathrm{Hopital}\right]$$$$\ln \mathcal{L}=\underset{x\to 0}{\lim }\frac{1}{\cos \sqrt{\mathrm{x}}}.\frac{\sin \sqrt{\mathrm{x}}}{2\sqrt{\mathrm{x}}}$$$$\underset{―}{\mathbf{ln}\mathcal{L}=1.\frac{1}{2}\iff \mathcal{L}={\mathbf{e}}^{\frac{1}{2}}=\sqrt{\mathbf{e}}}$$

Answered by mnjuly1970 last updated on 14/Jan/21

$$l={lim}_{x\to 0^{+}}\frac{1}{{\left[{(1-{sin}^2\left(\sqrt{x}\right))}^{\frac{1}{x}}\right]}^{\frac{1}{2}}}$$$$\underset{sin\left(\sqrt{x}\right)\approx \sqrt{x}}{\stackrel{x\to 0^{+}}{=}}{lim}_{x\to 0^{+}}{\left(\frac{1}{{(1-x)}^{\frac{1}{x}}}\right)}^{\frac{1}{2}}=\sqrt{e}...✓✓$$

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