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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 familia$r$ with this site too much, how do you accept answers? *edit nevermind i see now2011-07-01

2 Answers 2

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$\frac{1}{x} \rightarrow \infty$ 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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    @Christian: Then I shall edit it to your favor presently.2011-06-26