Question

The value of $\underset{x\to 0}{lim}\frac{logcox}{x^2}$ using taylor series is?

• A. $4$
• B. $\frac{2}{3}$
• C. $\frac{-1}{2}$
• D. $-2$

Answer 1

The correct option is C $\frac{-1}{2}$

Here, $f\left(x\right)=log\left(cos\right(x\left)\right)$
$f^{\prime }\left(x\right)=-tan\left(x\right)$ and $f^{\prime }\left(0\right)=0$
$f^{\prime \prime }\left(x\right)=-ta{n}^2\left(x\right)-1$ and $f^{\prime \prime }\left(0\right)=-1$
$f^{\prime \prime \prime }\left(x\right)=-2tan\left(x\right)\left(ta{n}^2\right(x)+1)$ and $f^{\prime \prime \prime }\left(0\right)=0$
$f^{\prime \prime \prime \prime }\left(x\right)=-6ta{n}^4\left(x\right)-8ta{n}^2\left(x\right)$ and $f^{\prime \prime \prime \prime }\left(0\right)=-2$
So,
$f\left(x\right)\approx \frac{0}{0!}{x}^0+\frac{0}{1!}{x}^1-\frac{1}{2!}{x}^2+\frac{0}{3!}{x}^4-\frac{2}{4!}{x}^4+........$
Therefore,
$log\left(cos\right(x\left)\right)\approx -\frac{1}{2}{x}^2-\frac{1}{12}{x}^4$
Divide by $x^2$ on both the sides, we get
$\frac{log\left(cos\right(x\left)\right)}{x^2}=-\frac{1}{2}-\frac{1}{12}{x}^2$
${lim}_{x\to 0}\frac{log\left(cos\right(x\left)\right)}{x^2}={lim}_{x\to 0}(-\frac{1}{2}-\frac{1}{12}{x}^2)=-\frac{1}{2}$