How do you use l'Hopital's Rule to evalute lim_(xrarroo) (x+sin(x))/x?

Oct 18, 2016

L'Hopital's Rule only applies if the limit is the indeterminate $\frac{0}{0}$

Explanation:

${\lim}_{x \rightarrow 00} \frac{x \cdot \sin \left(x\right)}{x} \ne {\lim}_{x \rightarrow \infty} \frac{0}{0}$

Oct 18, 2016

There is one error and an additional explanation is needed for why it won't work even if we correct the error. I have added a second edit .

Explanation:

Error

$\frac{d}{\mathrm{dx}} \left(x \sin x\right) = \sin x + x \cos x$ $\text{ }$ (Use the product rule.)

Corrected version

${\lim}_{x \rightarrow \infty} \frac{x \sin x}{x}$ has initial form $\frac{\infty}{\infty}$.

We apply l"Hopital's Rule for $f \frac{x}{g} \left(x\right)$ and attempt to find

${\lim}_{x \rightarrow \infty} \frac{\sin x + x \cos x}{1}$.

This limit does not exist and is neither $\infty$ nor $- \infty$, therefore l'Hopital does not apply.

Explanation

In order to use

${\lim}_{x \rightarrow a} f \frac{x}{g} \left(x\right) = {\lim}_{x \rightarrow a} f ' \frac{x}{g} ' \left(x\right)$

we must have one of the appropriate initial indeterminate forms ($\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$)

and also we must have

${\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ is finite or $\infty$ or $- \infty$.

In this question, ${\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ is not helpful.

Second Edit

It occurs to me that there may be a typographic error in the question.

It may be that the intended question involved $f \frac{x}{g} \left(x\right) = \frac{x + \sin x}{x}$.

Is so, then the derivative stated in the question is correct, but the explanation of why l'Hopital fails is the same.

In order to get an anwer from l'Hopital we must have

${\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ is finite or $\infty$ or $- \infty$..

But ${\lim}_{x \rightarrow \infty} \left(1 + \cos x\right)$ does not satisfy this condition.

Therefore l'Hopital's rule offers us no help for evaluating this limit.

(Note: This example shows the l'Hopital will not give answers for every indeterminate limit. Sometimes it fails.)

Finally, note that we can evaluate the limit without l'Hopital using $\frac{x + \sin x}{x} = 1 + \sin \frac{x}{x}$ which goes to $1 + 0 = 1$ as $x \rightarrow \infty$