Question

$\begin{array}{c}lim\\ x\to 1\end{array}\left[\frac{{\int }_0^{(x-1{)}^2}t\cos \left(t^2\right)dt}{(x-1)\sin (x-1)}\right]$

• A. is equal to $1$
• B. is equal to $\frac{1}{2}$
• C. does not exist
• D. is equal to $0$

Answer 1

The correct option is D is equal to $0$

$\begin{array}{c}lim\\ x\to 1\end{array}\left[\frac{{\int }_0^{(x-1{)}^2}t\cos \left(t^2\right)dt}{(x-1)\sin (x-1)}\right]$
Using L - Hospital rule
$=\begin{array}{c}lim\\ x\to 1\end{array}\frac{2(x-1).(x-1{)}^2\cos (x-1{)}^4-0}{(x-1)\cdot \cos (x-1)+\sin (x-1)}\left(\frac{0}{0}\right)$
$=\begin{array}{c}lim\\ x\to 1\end{array}\frac{2(x-1{)}^3\cdot \cos (x-1{)}^4}{(x-1)\left[\cos \right(x-1)+\frac{\sin (x-1)}{(x-1)}]}$
$\begin{array}{c}=lim\\ x\to 1\end{array}\frac{2(x-1{)}^2\cos (x-1{)}^4}{\cos (x-1)+\frac{\sin (x-1)}{(x-1)}}$
On taking limit
$=\frac{0}{1+1}=0$