Question

Solve the given limit problem $\underset{x\to 1}{\lim }\frac{{\int }_0^{{\left(x-1\right)}^2}t\cos \left(t^2\right)dt}{\left(x-1\right)\sin \left(x-1\right)}$.

• A. is equal to $1$
• B. is equal to $\frac{1}{2}$
• C. $0$
• D. is equal to $-\frac{1}{2}$

Answer 1

The correct option is C

$0$

Explanation for the correct option:

The given expression is,$\underset{x\to 1}{\lim }\frac{{\int }_0^{{\left(x-1\right)}^2}t\cos \left(t^2\right)dt}{\left(x-1\right)\sin \left(x-1\right)}$ $\left(1\right)$

Step-1: Finding the value of ${\int }_0^{{\left(x-1\right)}^2}t\cos \left(t^2\right)dt$

First we solve,${\int }_0^{{\left(x-1\right)}^2}t\cos \left(t^2\right)dt$

$$\begin{gathered} u=t^2\Rightarrow du=2t.dt \\ t\to 0\Rightarrow u=0 \\ t\to {\left(1-x\right)}^2\Rightarrow u={\left({\left(1-x\right)}^2\right)}^2={\left(1-x\right)}^4 \end{gathered}$$

$$\begin{gathered} {\int }_0^{{\left(x-1\right)}^4}\frac{\cos \left(u\right)}{2}du=\frac{1}{2}{\left[\sin \left(u\right)\right]}_0^{{\left(x-1\right)}^4} \\ =\frac{1}{2}\left[\sin {\left(x-1\right)}^4-\sin \left(0\right)\right] \\ =\frac{1}{2}\sin {\left(x-1\right)}^4\left[\because \sin \left(0\right)=0\right] \end{gathered}$$

On, substituting, Equation $\left(1\right)$ becomes

$\underset{x\to 1}{\lim }\frac{{\int }_0^{{\left(x-1\right)}^2}t\cos \left(t^2\right)dt}{\left(x-1\right)\sin \left(x-1\right)}=\underset{x\to 1}{\lim }\frac{\frac{1}{2}\sin {\left(x-1\right)}^4}{\left(x-1\right)\sin \left(x-1\right)}$

Step-2: Using L'hospitas rule

Now, at $x=1$, the numerator $\frac{1}{2}\sin {\left(x-1\right)}^4=\frac{1}{2}\sin {\left(1-1\right)}^4=0$ and the denominator $\left(x-1\right)\sin \left(x-1\right)=\left(1-1\right)\sin \left(1-1\right)=0$.

Thus, substituting the limit $x\to 1$, we get $\frac{0}{0}$ indeterminate form which means limit don't exist

So we can apply the L'hospitals rule which is $\underset{x\to 1}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 1}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

Step-3: Finding the limit

Now, applying the L'hospitals rule. we get:

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{\frac{1}{2}\sin {\left(x-1\right)}^4}{\left(x-1\right)\sin \left(x-1\right)}=\underset{x\to 1}{\lim }\frac{\frac{1}{2}\times \left[\cos {\left(x-1\right)}^4\right]\times 4{\left(x-1\right)}^3}{\left(x-1\right)\cos \left(x-1\right)+\sin \left(x-1\right)} \\ =\underset{x\to 1}{\lim }\frac{2{\left(x-1\right)}^3\cos {\left(x-1\right)}^4}{\left(x-1\right)\left[\cos \left(x-1\right)+{\displaystyle \frac{\sin \left(x-1\right)}{x-1}}\right]} \\ =\underset{x\to 1}{\lim }\frac{2{\left(1-x\right)}^2\cos {\left(1-x\right)}^4}{\left[\cos \left(x-1\right)+{\displaystyle \frac{\sin \left(x-1\right)}{x-1}}\right]} \\ =\frac{2{\left(1-1\right)}^3\times \cos {\left(1-1\right)}^4}{\left[\cos \left(1-1\right)+{\displaystyle \underset{x\to 1}{\lim }\frac{\sin \left(x-1\right)}{x-1}}\right]} \\ =\frac{0}{\left[1+{\displaystyle 1}\right]}\mathbf{}\left[\because \underset{x\to 1}{\lim }\frac{\sin \left(x-1\right)}{x-1}=1;\cos \left(0\right)=1\right] \\ =0 \end{gathered}$$

So, Limit of given function exist

Hence, the correct answer is option (C).