Chapter 9.2, Problem 3QQ

To determine

To Find : Limiting value of $\underset{x\to 2}{\lim }\frac{{\displaystyle \underset{2}{\overset{x}{\int }}\sin tdt}}{x^2-4}$

Answer to Problem 3QQ

  $\underset{x\to 2}{\lim }\frac{{\displaystyle \underset{2}{\overset{x}{\int }}\sin tdt}}{x^2-4}=\frac{\sin 2}{4}$

Answer (B).

Explanation of Solution

Given information :

  $\underset{x\to 2}{\lim }\frac{{\displaystyle \underset{2}{\overset{x}{\int }}\sin tdt}}{x^2-4}$

Formula used : L’Hospital Rule.

Calculation :

  $\begin{array}{l}\underset{x\to 2}{\lim }\frac{{\displaystyle \underset{2}{\overset{x}{\int }}\sin tdt}}{x^2-4}\backslash \\ =\underset{x\to 2}{\lim }\frac{{\left[-\cos \right]}_2^x}{x^2-4}\\ =\underset{x\to 2}{\lim }\frac{\cos 2-\cos x}{x^2-1}\left[\equiv \frac{0}{0}\right]\\ =\underset{x\to 2}{\lim }\frac{\sin x}{2x}\\ =\frac{\sin 2}{4}\end{array}$

Therefore, $\underset{x\to 2}{\lim }\frac{{\displaystyle \underset{2}{\overset{x}{\int }}\sin tdt}}{x^2-4}=\frac{\sin 2}{4}$