# Simple question regarding L'Hopital's rule

I understand how to do the problem except the answer is confusing me

$$\lim_{t\rightarrow0} \frac{e^t-1}{t^3}$$

After taking the derivative I ended up with $\dfrac{e^t}{3t^2}$ and that would go to $\frac{1}{0}$ which doesn't seem like a reasonable answer. The answer book says it would be positive ∞. Why?

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not familiar with this site too much, how do you accept answers? *edit nevermind i see now – Ryan Jul 1 '11 at 1:15

$\frac{1}{x} \rightarrow \infty$ as $x \rightarrow 0^+$ and $\frac{1}{x} \rightarrow -\infty$ as $x \rightarrow 0^-$.

Since in your case the denominator is always positive, it approaches $0$ from the positive direction, so the limit is $\infty$.

As a comment to another answer points out, you should be careful about saying that $1/0 = \infty$: $$\lim_{x \rightarrow 0} \frac1x$$ is undefined because $$\lim_{x \rightarrow 0^+} \frac1x = \infty \neq -\infty = \lim_{x \rightarrow 0^-} \frac1x$$

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The limit approaching 1/0 actually approaches $\infty$. So you're both right.

After one application of L'Hopital's rule, you get $\lim _{t \to 0} \dfrac{e^t}{3t^2}$. Note that both the top and the bottom are positive for all nonzero t. And recall that when we evaluate a limit, we don't just evaluate the value of the function at that point, but look at the behavior as it gets nearer the point. Well, as $t \to 0$, the top goes to 1, and the bottom gets arbitrarily small. Therefore, the number gets bigger and bigger.

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Not necessarily true: a limit whose numerator approaches $1$ and denominator approaches $0$ can also approach $-\infty$, or it can have no limit. E.g., $\frac{1}{t}$ as $t\to 0$ has no limit, $\frac{1}{-t^2}$ has limit $-\infty$. In the instant case you're right, but there's more to it than simply "1/0". – Arturo Magidin Jun 26 '11 at 0:22
@Arturo: yes, I meant with respect to this question in particular. I really meant for him to consider the sequence, rather than the limit itself. But I give fuller answers to non-homework questions and better answers to posters who comment, accept, and upvote answers. Thank you though - you are absolutely and obviously correct. (Though you must think my calculus to be very mediocre ;p) – mixedmath Jun 26 '11 at 0:32
Not at all; I thought perhaps you were thinking about this particular problem rather than making a general statement (since you say "the limit" rather than "a limit"), but it's too easy to misinterpret what you wrote, hence the comment. – Arturo Magidin Jun 26 '11 at 0:38
I don't understand how the sentence "The limit approaching 1/0 actually approaches $\infty$" could make 4 votes (so far). – Christian Blatter Jun 26 '11 at 8:08
@Christian: Then I shall edit it to your favor presently. – mixedmath Jun 26 '11 at 15:31