If the function $f(x)$, defined below, is continuous everywhere, then $k$ equals $f(x)=\left\{\begin{array}{cc}\frac{2^x-1}{\sqrt{1+x}-1}, & -1 \leq x

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If the function $f(x)$, defined below, is continuous everywhere, then $k$ equals $f(x)=\left\{\begin{array}{cc}\frac{2^x-1}{\sqrt{1+x}-1}, & -1 \leq x < \infty \ k, & x=0\end{array}ight.$

Aicharya · Accepted Answer

Given, $f(x)=\left\{\begin{array}{ccc}\frac{2^x-1}{\sqrt{1+x}-1}, & -1 \leq x < \infty \ k, & x=0\end{array}ight.$ is continuous everywhere. Since, $f(x)$ is continuous everywhere $\Rightarrow \quad f(x)$ will be continuous at $x=0$ $\Rightarrow \quad \lim _{x ightarrow 0} f(x)=f(0)$ $\lim _{x ightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1}=k$ $\left[\frac{0}{0}ight.$ form, so by applying $L^{\prime}$ Hospital rule $]$ $\begin{aligned} & \lim _{x ightarrow 0} \frac{2^x \log 2}{\frac{1}{2 \sqrt{1+x}}}=k \ & \lim _{x ightarrow 0} 2^{x+1} \sqrt{1+x} \log 2=k\end{aligned}$ $\begin{array}{ll}\Rightarrow & 2 \sqrt{1+0} \log 2=k \Rightarrow k=2 \log _e 2 \ \Rightarrow & k=\log _e 4\end{array}$

If the function $f(x)$, defined below, is continuous everywhere, then $k$ equals…

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