Question

If $\underset{x\to 0}{lim}\frac{(e^{kx}-1)\sin kx}{x^2}=4$, then $k=?$

• A. $2$
• B. $-2$
• C. $\pm 2$
• D. $\pm 4$

Answer 1

The correct option is C

$\pm 2$

Explanation for the correct option.

Step 1: Simplify $\underset{x\to 0}{lim}\frac{(e^{kx}-1)\sin kx}{x^2}$

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{(e^{kx}-1)\sin kx}{x^2}& =& \underset{x\to 0}{lim}\frac{(e^{kx}-1)}{x}\times \frac{\sin kx}{x}\\ & =& \underset{x\to 0}{lim}\frac{(e^{kx}-1)}{x}\times \frac{\sin kx}{kx}\times k\left(\mathrm{dividing}\mathrm{and}\mathrm{multiplying}\mathrm{by}k\right)\\ & =& \underset{x\to 0}{lim}\frac{(e^{kx}-1)}{x}\times 1\times k;\left(by\frac{\sin kx}{kx}=1\right).....\left(1\right)\\ & & \end{array}$

Step 2: Solve $\underset{x\to 0}{lim}\frac{(e^{kx}-1)}{x}$.

$\frac{(e^0-1)}{0}=\frac{0}{0}$

As, $\frac{(e^0-1)}{0}$ is in $\frac{0}{0}$form, so we will apply L's Hospital Rule.

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{(e^{kx}-1)}{x}& =& \underset{x\to 0}{lim}\frac{k·e^{kx}-0}{1}\\ & =& k·e^0\\ & =& k...........\left(2\right)\end{array}$

Step 3: Find the value of $k$.

From step 1 and step 2, we get

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{(e^{kx}-1)\sin kx}{x^2}& =& k\times 1\times k\\ & =& k^2\end{array}$

It is given that,

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{(e^{kx}-1)\sin kx}{x^2}& =& 4\\ \Rightarrow k^2& =& 4\\ \Rightarrow k& =& \pm 2\end{array}$

Hence, option C is correct.