Question

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

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

Answer 1

The correct option is A $abc$

$I=\underset{x\to 0}{lim}ln\left(\frac{1}{t}{\int }_0^t{(1+a\sin bx)}^{c/x}dx\right)=ln\left(\frac{1}{t}{\int }_0^t\underset{x\to 0}{lim}{(1+a\sin bx)}^{c/x}dx\right)$
limit is of the form $1^{\infty }$
$I=ln\left(\frac{1}{t}{\int }_0^t{e}^{{lim}_{x\to 0}c\left(\frac{1+a\sin bx-1}{x}\right)}dx\right)=ln\left(\frac{1}{t}{\int }_0^t{e}^{abc}dx\right)=ln\left(e^{abc}\frac{1}{t}{\int }_0^{t}dx\right)=abc$
Ans: A