Question

\(\displaystyle\lim_{x\to 1}\left(\frac{\int_{0}^{(x-1)^2} t\cos\left(t^2\right)dt}{(x-1)\sin(x-1)}\right)\)

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

Hint:

Put \(u=x-1\); the numerator integrates exactly to \(\frac{1}{2}\sin(u^4)\), which is of order \(u^4\), while the denominator is of order \(u^2\).

Answer 1

The Correct Option is A

Let \(u=x-1\to 0\).
\(\int_0^{u^2} t\cos(t^2)\,dt=\left[\frac{\sin(t^2)}{2}\right]_0^{u^2}=\frac{\sin(u^4)}{2}\).
So the limit is \(\lim_{u\to0}\frac{\sin(u^4)}{2u\sin u}=\lim_{u\to0}\frac{\sin(u^4)}{u^4}\cdot\frac{u^4}{2u\sin u}=1\cdot\lim_{u\to 0}\frac{u^2}{2}\cdot\frac{u}{\sin u}=0\cdot\frac{1}{2}\cdot 1\).
The limit equals 0.
Answer: (1) is equal to 0