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Question Number 166522 by mnjuly1970 last updated on 21/Feb/22

Commented by CAIMAN last updated on 21/Feb/22

1

Answered by MJS_new last updated on 21/Feb/22

$$\underset{x\to 0}{\lim }(\frac{1}{\ln \cos x}+\frac{2}{{\sin }^{2}x})=\underset{x\to 0}{\lim }\frac{2\ln \cos x+{\sin }^{2}x}{{\sin }^{2}x\ln \cos x}=$$$$=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}(2\ln \cos x+{\sin }^{2}x)}{\frac{d}{dx}\left({\sin }^{2}x\ln \cos x\right)}=$$$$\left[\mathrm{transforming}\right]$$$$=\underset{x\to 0}{\lim }-\frac{{2\sin }^{2}x}{{2\cos }^{2}x\ln \cos x-{\sin }^{2}x}=$$$$=\underset{x\to 0}{\lim }-\frac{\frac{d}{dx}\left({2\sin }^{2}x\right)}{\frac{d}{dx}({2\cos }^{2}x\ln \cos x-{\sin }^{2}x)}=$$$$\left[\mathrm{transforming}\right]$$$$=\underset{x\to 0}{\lim }\frac{1}{1+\ln \cos x}=1$$

Commented by mnjuly1970 last updated on 24/Feb/22

$$thanksslotsir$$

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