Question

The value of $\mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\log }_e}\left( {{{\log }_e}x} \right)}}{{{e^{\sqrt x }}}}$ is
A. $3$
B. $2$
C. $1$
D. $0$

Answer 1

Hint : As we know that the above question is related to limit and its function . Limit is the value that a function or sequence approaches as the input also approaches the same value. In the question we have to find the value where $x$ tends to infinity, since we cannot get any value with infinity, so we will try to convert it into the finite term and solve the value.

Complete step by step solution:
Here we have $\mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\log }_e}\left( {{{\log }_e}x} \right)}}{{{e^{\sqrt x }}}}$ .
If we put infinity i.e. $\infty$ in the place of $x$ in numerator and denominator , then we get the value $\dfrac{\infty }{\infty }$ and there is no such value regarding it.
So we will apply the L’ Hopitals rule and solve it .
If we differentiate the ${\log _e}x$ , we get $\dfrac{1}{x}$ , and when we differentiate ${\log _e}$ , it gives the value $\dfrac{1}{{{{\log }_e}x}}$ .
We can write the numerator part as $\dfrac{1}{{x{{\log }_e}x}}$ .
Similarly when we differentiate $\sqrt x$ , then the value is $\dfrac{1}{{2\sqrt x }}$ and the value of $e$ is ${e^{\sqrt x }}$ .
By putting them all together we have
$\dfrac{1}{{\dfrac{{x{{\log }_e}x}}{{{e^{\sqrt x }}\dfrac{1}{{{2 \times {\sqrt x }}}}}}}}$ .
Now by simplifying and we have
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2\sqrt x }}{{x \times {{\log }_e}x \times {e^{\sqrt x }}}}$ .
On further solving we have
$\mathop {\lim }\limits_{x \to \infty } \dfrac{2}{{\sqrt x {{\log }_e}x \times {e^{\sqrt x }}}}$ . Now if we put the value of $\infty$ in place of $x$ , then the fraction is $\dfrac{2}{\infty }$ , finite term in the numerator and the value of $\dfrac{2}{\infty }$ is $0$ .
Hence the correct option is (d) $0$ .
So, the correct answer is “Option D”.

Note : Before solving this kind of question we should know all the rules and functions of limit and how to solve the problem. We know the formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ , So when we differentiate $\sqrt x$ , we can write it as $\dfrac{d}{{dx}}{x^{\dfrac{1}{2}}} = \dfrac{1}{2}{x^{\dfrac{{ - 1}}{2}}}$ as $\sqrt {}$ means the power $\dfrac{1}{2}$ . We have used the L’Hopital’s rule till we eliminate the indeterminate form like infinity.