Question

If $f(x)=\{\begin{array}{ll}\frac{e^{3x}-1}{4x}& \text{for}x\ne 0\\ \frac{k+x}{4}& \text{for }x=0\end{array}$ is continuous at $x=0$, then $k=$

• A. 5
• B. 3
• C. 2
• D. 0

Answer 1

Answer: (B)

If $f(x)=\{\begin{array}{ll}\frac{e^{3x}-1}{4x}& \text{for}x\ne 0\\ \frac{k+x}{4}& \text{for }x=0\end{array}$ is continuous at $x=0$,
$\therefore \underset{x\to 0+}{\lim }f\left(x\right)=\underset{x\to 0-}{\lim }f\left(x\right)=f\left(0\right)$
$\Rightarrow \underset{x\to 0+}{\lim }\frac{e^{3x}-1}{4x}=\frac{k}{4}$
$\Rightarrow \underset{x\to 0+}{\lim }\frac{e^{3h}-1}{4h}=\frac{k}{4}$ $\left(\frac{0}{0}from\right)$
Applying L-Hospital's Rule,
$\Rightarrow \underset{h\to 0}{\lim }\frac{3e^{3h}}{4}=\frac{k}{4}\Rightarrow k=3$