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Evaluating Limits using L'Hospital

shamieh

Active member
Sep 13, 2013
539
Need someone to check my work.


lim t -> 0

\(\displaystyle \frac{e^{2t} - 1}{1 - cos(t)}\)

after I took the derivative twice

I got \(\displaystyle \frac{2}{0}\) = undefined?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You can determine the limit does not exist after just one application of L'Hôpital's Rule. I would also look at the one-sided limits. What to you find?

Note: use the $\LaTeX$ code \lim_{t\to0} for your limit.
 

shamieh

Active member
Sep 13, 2013
539
Oh because the denominator has the cos and sin. that's what you are saying correct? How I could of known after 1 application of LHopsital...So is this the correct answer though? Does not exist? Or Should I put "Undefined"?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
This is what I would write:

\(\displaystyle L=\lim_{t\to0}\frac{e^{2t}-1}{1-\cos(t)}\)

This is the indeterminate form \(\displaystyle \frac{0}{0}\), so application of L'Hôpital's rule yields:

\(\displaystyle L=\lim_{t\to0}\frac{2e^{2t}}{\sin(t)}=\frac{2}{0}\)

This is undefined, so we want to look at the one-sided limits:

\(\displaystyle \lim_{t\to0^{-}}\frac{2e^{2t}}{\sin(t)}=-\infty\)

\(\displaystyle \lim_{t\to0^{+}}\frac{2e^{2t}}{\sin(t)}=\infty\)

Since:

\(\displaystyle \lim_{t\to0^{-}}\frac{2e^{2t}}{\sin(t)}\ne \lim_{t\to0^{+}}\frac{2e^{2t}}{\sin(t)}\) we may conclude that the limit $L$ does not exist.