Question

If $f(x)=\{\begin{array}{ll}\log {\left({\sec }^{2}x\right)}^{{\cot }^{2}x},& \text{ for }x\ne 0\\ K,& x=0\end{array}$
is continuous at $x=0$ then $K$ is

• A. $e^{-1}$
• B. $1$
• C. $e$
• D. $0$

Answer 1

Answer: (B)

Since, $f$ is continuous at $x=0$
$\therefore f(0)=\underset{x\to 0}{\lim }\log {\left({\sec }^{2}x\right)}^{{\cot }^{2}x}$
Therefore we have,
$K=\underset{x\to 0}{\lim }{\cot }^{2}x\cdot \log \left({\sec }^{2}x\right)$
$=\underset{x\to 0}{\lim }{\cot }^{2}x\cdot \log \left(1+{\tan }^{2}x\right)$
$=\underset{x\to 0}{\lim }\frac{\log \left(1+{\tan }^{2}x\right)}{{\tan }^{2}x}$
$\therefore K=1\dots \dots .\underset{x\to 0}{\lim }\frac{\log (1+x)}{x}=1$