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Compute \mathrm{ \lim _{x \rightarrow 0} \frac{\cos x-1}{\ln \left(1+\sin ^2 x\right)}}

Option: 1

-\frac{1}{2}


Option: 2

\frac{2}{3}


Option: 3

\frac{1}{4}


Option: 4

0


Answers (1)

best_answer

Using the series forms:
\mathrm{ \cos (x)=1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\cdots, \forall x }
and
\mathrm{ \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots, x \in(-1,1] }
then
\mathrm{ \lim _{x \rightarrow 0} \frac{\cos (x)-1}{\ln \left(1+\sin ^2(x)\right)} =\lim _{x \rightarrow 0} \frac{\left[1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\cdots\right]-1}{\sin ^2(x)-\frac{\left[\left.\sin ^2(x)\right|^2\right.}{2}+\frac{\left[\sin ^2(x)\right]^3}{3}-\cdots} }

\mathrm{ =\lim _{x \rightarrow 0} \frac{-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\cdots}{\sin ^2(x)\left[1-\frac{\sin ^2(x)}{2}+\frac{\sin ^4(x)}{3}-\cdots\right]} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} }

\mathrm{ =\lim _{x \rightarrow 0} \frac{-\frac{1}{2}+\frac{x^2}{4}-\cdots}{\underbrace{\left(\frac{\sin (x)}{x}\right)^2}_{\rightarrow 1}[1-\underbrace{\left.\frac{\sin ^2(x)}{2}+\frac{\sin ^4(x)}{3}-\cdots\right]}_{\rightarrow 0}]}=\frac{-\frac{1}{2}+0}{1 \cdot 1}=-\frac{1}{2}}

Hence option 1 is correct.



 

Posted by

Pankaj Sanodiya

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