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If $f(x)=\left\{\begin{array}{ll}\frac{e^{x}-1}{4 x} & \text { for } x \neq 0 \\ \frac{k+x}{4} & \text { for } x=0\end{array}\right.$ is continuous at $x=0$, then $k=$
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The correct answer is:
3
$\because f(x)$ is continuous at $x=0$
$$
\begin{aligned}
&\text { So, } \lim _{x \rightarrow 0} f(x)=f(0) \\
&\Rightarrow \quad \lim _{x \rightarrow 0} \frac{e^{3 x-1}}{4 x}=\frac{k+0}{4} \\
&\Rightarrow \quad \frac{1}{4} \lim _{x \rightarrow 0} 3\left(\frac{e^{3 x}-1}{3 x}\right)=\frac{\bar{k}}{4} \\
&\Rightarrow \quad 3 \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{3 x}=k \Rightarrow 3 \times 1=k \Rightarrow k=3
\end{aligned}
$$
$$
\begin{aligned}
&\text { So, } \lim _{x \rightarrow 0} f(x)=f(0) \\
&\Rightarrow \quad \lim _{x \rightarrow 0} \frac{e^{3 x-1}}{4 x}=\frac{k+0}{4} \\
&\Rightarrow \quad \frac{1}{4} \lim _{x \rightarrow 0} 3\left(\frac{e^{3 x}-1}{3 x}\right)=\frac{\bar{k}}{4} \\
&\Rightarrow \quad 3 \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{3 x}=k \Rightarrow 3 \times 1=k \Rightarrow k=3
\end{aligned}
$$
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