Question

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

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

Answer 1

The correct option is C $\pm 2$

$\underset{x\to 0}{lim}\frac{(e^{kx}-1)sinkx}{x^2}=4$

$\Rightarrow \frac{(e^{kx}-1)}{kx}\frac{sinkx}{kx}.k.k$

$\Rightarrow x\to 0\therefore kx\to 0$

$\underset{x\to 0}{lim}\frac{e^{kx}-1}{kx}=1;\underset{x\to 0}{lim}\frac{sinkx}{kx}=1$

$\therefore \underset{x\to 0}{lim}\frac{(e^{kx}-1)sinkx}{x^2}=k^2$

$\therefore k^2=4$

$k=\pm 2$