Question

Find the limit. Use L’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If L’Hospital’s Rule doesn’t apply, explain why.

19. \(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{{1 + {e^x}}}\)

Short answer

The value is \(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{{1 + {e^x}}} = 0\).

Step-by-step solution

L’Hospital’s Rule

TheL’Hospital’s Rulestates that for any real number\(a\), in case:

\(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{0}{0},{\rm{ or, }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{\infty }{\infty }\),

Then the limit of the function can be calculated as:

\(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\)

Solving for given limit.

The given limit is:

\(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{{1 + {e^x}}}\)

Here, we have \(f\left( x \right){\rm{ and }}g\left( x \right)\) both the functions are differentiable.

\(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{{1 + {e^x}}} = \frac{\infty }{\infty }\)

So, usingL’Hospital’s Rule, we have:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{{2\sqrt x }}}}{{{e^x}}}\\ = \frac{1}{\infty } = 0\end{array}\)

Hence, the required value is \(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{{1 + {e^x}}} = 0\).