Question

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

• A. 5
• B. 3
• C. 2
• D. 0

Answer 1

Answer: (B)

3

Answer 2

If $f(x) = \begin{cases} \frac{e^{3x} - 1}{4x} & \quad \text{for} x \neq 0 \\\ \frac{k + x}{4} & \quad \text{for } x= 0 \end{cases}$ is continuous at $x = 0$,
$\therefore \:\:\: \displaystyle\lim_{x\to0+}f\left(x\right) = \displaystyle\lim _{x\to 0-}f\left(x\right) =f\left(0\right)$
$\Rightarrow \displaystyle\lim _{x\to 0+} \frac{e^{3x} - 1}{4x} = \frac{k}{4}$
$\Rightarrow \displaystyle\lim _{x\to 0+} \frac{e^{3h} - 1}{4h} = \frac{k}{4}$ $\left( \frac{0}{0} \, from \right)$
Applying L-Hospital's Rule,
$\Rightarrow \displaystyle\lim_{h \rightarrow0} \frac{3e^{3h}}{4} = \frac{k}{4} \Rightarrow k=3$