Question

Evaluate $\underset{x\to 0}{\lim }\frac{{\int }_0^{x}t\sin \left(10t\right)dt}{x}$ is equal to

• A. $0$
• B. $\frac{1}{10}$
• C. $-\frac{1}{10}$
• D. $-\frac{1}{5}$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Find the value of the given Equation.

Consider the given Equation as

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{{\int }_0^{x}t\sin \left(10t\right)dt}{x} \\ I=\frac{{\int }_0^{x}0\sin \left(10\left(0\right)\right)dt}{0} \\ I=\frac{0}{0}\left[\text{Indeterminent form}\right] \end{gathered}$$

Using the L' hospital rule

$$\begin{gathered} I=\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)} \\ =\underset{x\to c}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)} \end{gathered}$$

Then,

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{{\int }_0^{x}t\sin \left(10t\right)dt}{x} \\ I=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}\left({\int }_0^{x}t\sin \left(10t\right)dt\right)}{\frac{d}{dx}\left(x\right)}\left[\because \underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}\right] \\ I=\underset{x\to 0}{\lim }\frac{{\left[t\sin \left(10t\right)\right]}_0^x}{1}\left[\because \frac{d}{dx}\int f\left(x\right)dx=f\left(x\right)\right] \\ I=\underset{x\to 0}{\lim }\frac{x\sin \left(10x\right)-0}{1} \\ I=\underset{x\to 0}{\lim }\frac{\left(0\right)\sin \left(10\left(0\right)\right)}{1} \\ I=\frac{0}{1} \\ I=0 \end{gathered}$$

Hence, option (A) is the correct answer.