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
![lim_(x→0) ((1/(ln cos x))+(2/(sin^2 x))) =lim_(x→0) ((2ln cos x +sin^2 x)/(sin^2 x ln cos x)) = =lim_(x→0) (((d/dx)(2ln cos x +sin^2 x))/((d/dx)(sin^2 x ln cos x))) = [transforming] =lim_(x→0) −((2sin^2 x)/(2cos^2 x ln cos x −sin^2 x)) = =lim_(x→0) −(((d/dx)(2sin^2 x))/((d/dx)(2cos^2 x ln cos x −sin^2 x))) = [transforming] =lim_(x→0) (1/(1+ln cos x)) =1](https://www.tinkutara.com/question/Q166527.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{cos}\:{x}}+\frac{\mathrm{2}}{\mathrm{sin}^{\mathrm{2}} \:{x}}\right)\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2ln}\:\mathrm{cos}\:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{ln}\:\mathrm{cos}\:{x}}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left(\mathrm{2ln}\:\mathrm{cos}\:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}\right)}{\frac{{d}}{{dx}}\left(\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{ln}\:\mathrm{cos}\:{x}\right)}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{transforming}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:−\frac{\mathrm{2sin}^{\mathrm{2}} \:{x}}{\mathrm{2cos}^{\mathrm{2}} \:{x}\:\mathrm{ln}\:\mathrm{cos}\:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:−\frac{\frac{{d}}{{dx}}\left(\mathrm{2sin}^{\mathrm{2}} \:{x}\right)}{\frac{{d}}{{dx}}\left(\mathrm{2cos}^{\mathrm{2}} \:{x}\:\mathrm{ln}\:\mathrm{cos}\:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}\right)}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{transforming}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{ln}\:\mathrm{cos}\:{x}}\:=\mathrm{1} \\ $$
Commented by mnjuly1970 last updated on 24/Feb/22
$${thanks}\:{slot}\:{sir} \\ $$