If the function f(x),defined below,is continu | vedclass

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(B) Given,$f(x) = \begin{cases} \frac{2^x-1}{\sqrt{1+x}-1}, & x 
eq 0 \ k, & x=0 \end{cases}$ is continuous everywhere. Since $f(x)$ is continuous e

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$\log _e 4$

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(B) Given,$f(x) = \begin{cases} \frac{2^x-1}{\sqrt{1+x}-1}, & x 
eq 0 \ k, & x=0 \end{cases}$ is continuous everywhere. Since $f(x)$ is continuous everywhere,it must be continuous at $x=0$. Therefore,$\lim_{x ightarrow 0} f(x) = f(0)$. $\lim_{x ightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1} = k$. This is a $\frac{0}{0}$ form,so we apply $L^{\prime}$ Hospital rule: $\lim_{x ightarrow 0} \frac{\frac{d}{dx}(2^x-1)}{\frac{d}{dx}(\sqrt{1+x}-1)} = k$. $\lim_{x ightarrow 0} \frac{2^x \log_e 2}{\frac{1}{2\sqrt{1+x}}} = k$. Substituting $x=0$: $\frac{2^0 \log_e 2}{\frac{1}{2\sqrt{1+0}}} = k$. $\frac{1 \cdot \log_e 2}{\frac{1}{2}} = k$. $2 \log_e 2 = k$. Using the property $n \log a = \log a^n$,we get $k = \log_e 2^2 = \log_e 4$.

If the function $f(x)$,defined below,is continuous everywhere,then $k$ equals: $f(x)=\begin{cases} \frac{2^x-1}{\sqrt{1+x}-1}, & x \neq 0 \\ k, & x=0 \end{cases}$

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