If f(x) = log(sec^2 x)^{cot^2 x} for x neq 0 | vedclass

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(C) For a function to be continuous at $x=0$,the condition $\lim_{x \rightarrow 0} f(x) = f(0)$ must hold.Given $f(0) = K+1$.We evaluate the limit: $L

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(C) For a function to be continuous at $x=0$,the condition $\lim_{x ightarrow 0} f(x) = f(0)$ must hold.Given $f(0) = K+1$.We evaluate the limit: $L = \lim_{x ightarrow 0} \log(\sec^2 x)^{\cot^2 x}$.Using the property of logarithms,$L = \lim_{x ightarrow 0} \cot^2 x \cdot \log(\sec^2 x)$.Since $\sec^2 x = 1 + 	an^2 x$,we have $L = \lim_{x ightarrow 0} \frac{\log(1 + 	an^2 x)}{	an^2 x}$.Let $u = 	an^2 x$. As $x ightarrow 0$,$u ightarrow 0$. The limit becomes $\lim_{u ightarrow 0} \frac{\log(1+u)}{u}$.Using the standard limit $\lim_{u ightarrow 0} \frac{\log(1+u)}{u} = 1$,we get $L = 1$.Equating the limit to $f(0)$,we have $1 = K+1$.Therefore,$K = 0$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

If $f(x) = \log(\sec^2 x)^{\cot^2 x}$ for $x \neq 0$ and $f(x) = K+1$ for $x=0$ is continuous at $x=0$,then the value of $K$ is

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