Question

$\underset{x\to 1}{\lim }\left(\frac{\underset{0}{\overset{(x-1{)}^2}{\int }}t\cos \left(t^2\right)dt}{(x-1)\sin (x-1)}\right)$

• A. does not exist
• B. is equal to $\frac{1}{2}$
• C. is equal to 1
• D. None of the above

Answer 1

Answer: (D)

$\underset{x\to 1}{\lim }\frac{\underset{0}{\overset{(x-1{)}^2}{\int }}tcos(t^2)dt}{(x-1)\sin (x-1)}\left(\frac{0}{0}\right)$
Apply L Hopital Rule
$=\underset{x\to 1}{\lim }\frac{2(x-1)\cdot (x-1{)}^2\cos (x-1{)}^4-0}{(x-1)\cdot \cos (x-1)+\sin (x-1)}\left(\frac{0}{0}\right)$
$=\underset{x\to 1}{\lim }\frac{2(x-1{)}^3\cdot \cos (x-1{)}^4}{(x-1)\left[\cos (x-1)+\frac{\sin (x-1)}{(x-1)}\right]}$
$=\underset{x\to 1}{\lim }\frac{2(x-1{)}^2\cos (x-1{)}^4}{\cos (x-1)+\frac{\sin (x-1)}{(x-1)}]}$
$=\underset{x\to 1}{\lim }\frac{2(x-1{)}^2\cos (x-1{)}^4}{\cos (x-1)+\frac{\sin (x-1)}{(x-1)}}$
on taking limit
$=\frac{0}{1+1}=0$