Question

If $a,b$ and $c$ are real numbers then the value of $\underset{t\to 0}{lim}{l}_n\left(\frac{1}{t}{\int }_0^1{\left(1+a\sin bx\right)}^{\frac{c}{x}}dx\right)$ equals

• A. $abc$
• B. $\frac{ab}{c}$
• C. $\frac{bc}{a}$
• D. $\frac{ca}{b}$

Answer 1

The correct option is A $abc$

Let $L={lim}_{t\to 0}\ln \left(\frac{1}{t}{\int }_0^t{(1+a\sin bx)}^{\frac{c}{x}}dx\right)$

It is of $\frac{0}{0}$ form.

Hence we use L'Hospitals's rule

$L={lim}_{t\to 0}\frac{{(1+a\sin bt)}^{\frac{c}{t}}\times 1}{1}$

$L={lim}_{t\to 0}\ln e^{a\sin bt\times \frac{c}{t}}$

$L={lim}_{t\to 0}\frac{ac\sin bt}{t}$ is of $\frac{0}{0}$ form.

$L={lim}_{t\to 0}\frac{ac\cos bt}{1}$

$=abc$