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### Section 4-10 : L'Hospital's Rule and Indeterminate Forms

11. Use L’Hospital’s Rule to evaluate $$\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}}$$.

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The first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on certain classes of rational functions and this is clearly not a rational function.

Note however that it is in the following indeterminate form,

${\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}{\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}} \to {\infty ^0}$

and as we discussed in the notes for this section we can do some manipulation on this to turn it into a problem that can be done with L’Hospital’s Rule.

Show Step 2

First, let’s define,

$z = {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}}$

and take the log of both sides. We’ll also do a little simplification.

$\ln z = \ln \left( {{{\left[ {{{\bf{e}}^x} + x} \right]}^{{}^{1}/{}_{x}}}} \right) = \frac{1}{x}\ln \left[ {{{\bf{e}}^x} + x} \right] = \frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}$ Show Step 3

We can now take the limit as $$x \to \infty$$ of this.

$\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}} \right]$

Before we proceed let’s notice that we have the following,

${\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x} \to = \frac{\infty }{\infty }$

and we have a limit that we can use L’Hospital’s Rule on.

Show Step 4

Applying L’Hospital’s Rule gives,

$\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{{}^{{{{\bf{e}}^x} + 1}}/{}_{{{{\bf{e}}^x} + x}}}}{1} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}}$ Show Step 5

We now have a limit that behaves like,

${\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}\frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}} \to \frac{\infty }{\infty }$

and so we can use L’Hospital’s Rule on this as well. Doing this gives,

$\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x}}}{{{{\bf{e}}^x} + 1}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x}}}{{{{\bf{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \left( 1 \right) = 1$

Notice that we did have to use L’Hospital’s Rule twice here and we also made sure to do some simplification so we could actually take the limit.

Show Step 6

Now all we need to do is recall that,

$z = {{\bf{e}}^{\ln z}}$

This in turn means that we can do the original limit as follows,

$\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}} = \mathop {\lim }\limits_{x \to \infty } z = \mathop {\lim }\limits_{x \to \infty } {{\bf{e}}^{\ln z}} = {{\bf{e}}^{\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right]}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\bf{e}}}$