Question

The value of $\underset{x\to 0}{\lim }\frac{\left[{\int }_0^x\cos \left(t\right)dt\right]}{x}$ is

• A. $1$
• B. $-1$
• C. $\infty$
• D. None of these.

Answer 1

The correct option is A

$1$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 0}{\lim }\frac{\left[{\int }_0^x\cos \left(t\right)dt\right]}{x}$

Using Leibnitz's rule,

$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{\sin \left(x\right)-\sin \left(0\right)}{x} \\ =\underset{x\to 0}{\lim }\frac{\sin \left(x\right)}{x} \\ =\frac{0}{0} \end{gathered}$$

Since it's $\frac{0}{0}$ form apply L.Hospital rule,

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\cos \left(x\right)}{1} \\ =\frac{1}{1} \\ =1 \end{gathered}$$

Therefore, the value of $\underset{x\to 0}{\lim }\frac{\left[{\int }_0^x\cos \left(t\right)dt\right]}{x}$ is $1$.

Hence option $\left(A\right)$ is the correct option.