Question

\[{\lim _{x \to 0}}\dfrac{{\int_0^x {t\sin (10t)dt} }}{x}\] is equal to:
(A) \[0\]
(B) \[\dfrac{1}{{10}}\]
(C) \[\dfrac{{ - 1}}{{10}}\]
(D) \[\dfrac{{ - 1}}{5}\]

Answer 1

Hint: TWhen we apply limit to the given question, we find that it is of type \[\dfrac{0}{0}\]. Hence, we can use L’Hopital’s rule to evaluate the limit. In L’Hopital’s rule, we differentiate the numerator and the denominator and then take the limit.

Formula Used:
According to L’Hopital’s rule if f and g are differentiable functions such that \[\mathop {\lim }\limits_{x \to a} f\left( x \right) = 0\] and \[\mathop {\lim }\limits_{x \to a} g\left( x \right) = 0\]then \[\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}\].

Complete step by step Solution:
We are given that
\[{\lim _{x \to 0}}\dfrac{{\int_0^x {t\sin (10t)dt} }}{x}\]
We observe that if we put x=0 in the given limit, we would get \[\dfrac{0}{0}\]. Therefore, the given limit is of \[\dfrac{0}{0}\] indeterminate form.
Hence, we must apply L – Hospital’s Rule to evaluate the limit.
The differentiation of\[\int_0^x {t\sin (10t)dt} \] is \[x\sin (10x)\] and the differentiation of \[x\]is \[1\].
Therefore, we have,
\[{\lim _{x \to 0}}\dfrac{{x\sin (10x)}}{1} = \dfrac{0}{1} = 0\] [Since \[\sin (10x) = 0\]]
Hence \[{\lim _{x \to 0}}\dfrac{{\int_0^x {t\sin (10t)dt} }}{x} = 0\]

Hence, the correct option is A.

Note: In order to solve the given question, one must know about the L’Hopital’s rule. L’Hopital’s rule is a method of evaluating indeterminate forms such as \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\]. To evaluate the limits of indeterminate forms for the derivatives in calculus, L’Hopital’s rule is used.