If $f(x)=\frac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^2}$ for $
eq \pi$, is continuous at $x=\pi$, then $=\mathrm{k} \quad$ for $=\pi$ $\mathrm{k}=$

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Aicharya · Accepted Answer

$f(\pi)=\lim _{x ightarrow \pi} f(x)=\lim _{x ightarrow \pi} \frac{4^{x-\pi}+4^{\pi-4}-2}{(x-\pi)^2}$ Put $\mathrm{x}=\pi+\mathrm{h}$. As $\mathrm{x} ightarrow \pi, \mathrm{h} ightarrow 0$ $\begin{aligned} & 	herefore \mathrm{f}(\pi)=\lim _{\mathrm{h} ightarrow 0} \frac{4^{\mathrm{h}}+4^{-\mathrm{h}}-1-1}{\mathrm{~h}^2}=\lim _{\mathrm{h} ightarrow 0} \frac{\left(4^{\mathrm{h}}-1ight)}{\mathrm{h}} 	imes \lim _{\mathrm{h} ightarrow 0} \frac{4^{-\mathrm{h}}-1}{-\mathrm{h}} 	imes(-1) \ & =(\log 4)(-\log 4)=(2 \log 2)(-2 \log 2)=-4(\log 2)^2 \end{aligned}$

If $f(x)=\frac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^2}$ for $\neq \pi$, is continuous at $x=\pi$, then…

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