Help with limit and l'Hospital rule: (e^t - 1)/(t^3)

[degreeplus]

\(\displaystyle \
\lim \frac{{e^t - 1}}{{t^3 }}
\\)

I am not sure how to approach this problem. Here is how I see it I use l'Hospital's rule by taking the derivative of the top and bottom until the variable t is a 6:

\(\displaystyle \
{\lim }\limits_{t \to 0} \frac{{e^t - 1}}{{t^3 }} = {\lim }\limits_{t \to 0} \frac{{e^t }}{{3t^2 }} = {\lim }\limits_{t \to 0} \frac{{e^t }}{{6t}} = {\lim }\limits_{t \to 0} \frac{{e^t }}{6} = \frac{{e^0 }}{6} = \frac{1}{6}
\\)

Basically the answer is suppose to turn out to be

\(\displaystyle \
\infty
\\)

But I don't understand how to calculate it. I can see based off the graph of

\(\displaystyle \
\frac{{e^t - 1}}{{t^3 }}
\\)

but have trouble figuring it out. Thanks for any help in advance.

 

[galactus]

You just went a little too far.

\(\displaystyle \L\\\lim_{t\to\0}\frac{e^{t}-1}{t^{3}}\)

L'Hopital:

\(\displaystyle \L\\\frac{1}{3}\lim_{t\to\0}\frac{e^{t}}{t^{2}}\)

\(\displaystyle \L\\\frac{1}{3}\underbrace{\lim_{t\to\0}e^{t}}_{\text{limit=1}}\cdot\underbrace{\lim_{t\to\0}\frac{1}{t^{2}}}_{\text{limit=infinity}}\)

\(\displaystyle \L\\={\infty}\)

 

[degreeplus]

But that still does not make sense to me.

I don't see how the
\(\displaystyle \
{\lim }\limits_{t \to 0} \frac{1}{{t^2 }}\\) = \(\displaystyle \infty \\)

Isn't \(\displaystyle \frac{1}{{0^2 }}\) indeterminant ?

The only way I see it is by looking at the graph.

 

[stapel]

Isn't 1/0 indeterminant ?

Zero over zero, or 1 to the power zero, etc, are indeterminate, but 1/0 is one of those limits you should have learned near the beginning:

If the numerator is fixed and finite (in this case, at 1) and the denominator gets increasingly small, getting as close to zero as you could ever want, then the limit is undefined. Informally, we say "the limit goes to infinity", though of course "infinity" is not a number that one can "go" to.... :wink:

Eliz.

 

[galactus]

I am glad Stapel pointed that out. We say it is unbounded or the limit does not exist because it does not approach a fixed value.

As t approaches 0, 1/t^3 gets larger and larger. We don't actually get to 0, though.

 

[tkhunny]

Let me stress that you CANNOT apply l'Hospital's Rule willy-nilly. It applies ONLY to indeterminate forms. When you get to \(\displaystyle \frac{e^{t}}{3t^{2}}\), you cannot use it any more. It is not an indeterminate form.