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.

26. \(\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{{{u^3}}}\)

Short answer

The limit is \(\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{{{u^3}}} = \infty \).

Step-by-step solution

L’Hospital’s Rule

The L’Hospital’s Rule states that for any real number \(a\), in case:

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

Then the limit of the function can be calculated as:

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

Solving for given limit.

The given limit is:

\(\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{{{u^3}}}\)

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

\(\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{{{u^3}}} = \frac{\infty }{\infty }\)

So, usingL’Hospital’s Rule, we have:

\(\begin{array}{c}\mathop {\lim }\limits_{u \to \infty } \frac{{f\left( u \right)}}{{g\left( u \right)}} = \mathop {\lim }\limits_{u \to \infty } \frac{{f'\left( u \right)}}{{g'\left( u \right)}}\\ = \mathop {\lim }\limits_{u \to \infty } \frac{{\frac{1}{{10}}\left( {{e^{\frac{u}{{10}}}}} \right)}}{{3{u^2}}}\\ = \frac{1}{{30}}\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{{{u^2}}}\end{array}\)

Again, usingL’Hospital’s Ruletwo times, we have:

\(\begin{array}{c}\frac{1}{{30}}\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{{{u^2}}} = \frac{1}{{600}}\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{u}\\ = \frac{1}{{6000}}\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{1}\\ = \frac{{{e^\infty }}}{{6000}}\\ = \infty \end{array}\)

Hence, the required value is \(\mathop {\lim }\limits_{u \to \infty } \frac{{{e^{\frac{u}{{10}}}}}}{{{u^3}}} = \infty \).