integragirl
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- Apr 13, 2006
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Please help me as soon as possible
lim as t-->0 t^3/tan^3(2t)
lim as t-->0 t^3/tan^3(2t)
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Please help! lim as t-->0 t^3/tan^3(2t)
- Thread starter integragirl
- Start date
Since the limit of the numerator is 0, and the limit of the denominator is also 0, this is an indeterminate form of type 0/0. To solve this, you can use L'Hospitals Rule (take the derivative of the numerator and the dominator separately and re-evaluate the limit).
Remember, you can use that rule as many times as necessary as long as the limit is of the form 0/0 or oo/oo
Remember, you can use that rule as many times as necessary as long as the limit is of the form 0/0 or oo/oo
Hello, integragirl!
This looks like a pre-L'Hopital problem . . .
Recall that: \(\displaystyle \L\,\lim_{\theta\to0}\,\frac{\sin\theta}{\theta}\;=\;1\)
We have: \(\displaystyle \L\,\frac{t^3}{\frac{\sin^3(2t)}{\cos^3(2t)}} \;=\;\frac{t^3\cdot\cos^3(2t)}{\sin^3(2t)}\)
Multiply by \(\displaystyle \frac{8}{8}:\L\;\;\frac{8}{8}\,\cdot\frac{t^3\cdot\cos^3(2t)}{\sin^3(2t)} \;= \;\frac{1}{8}\cdot\cos^3(2t)\,\cdot\,\L\frac{8t^3}{\sin^3(2t)}\;=\;\frac{1}{8}\cdot\cos^3(2t)\,\cdot\,\L\left(\frac{2t}{\sin(2t)}\right)^3\)
Then: \(\displaystyle \L\,\lim_{2t\to0}\,\left[\frac{1}{8}\cdot\cos^3(2t)\cdot\left(\frac{2t}{\sin(2t)}\right)^3\right] \;=\;\frac{1}{8}\cdot1^3\cdot1^3\;=\;\frac{1}{8}\)
This looks like a pre-L'Hopital problem . . .
Recall that: \(\displaystyle \L\,\lim_{\theta\to0}\,\frac{\sin\theta}{\theta}\;=\;1\)
\(\displaystyle \L\lim_{t\to0}\, \frac{t^3}{\tan^3(2t)}\)
We have: \(\displaystyle \L\,\frac{t^3}{\frac{\sin^3(2t)}{\cos^3(2t)}} \;=\;\frac{t^3\cdot\cos^3(2t)}{\sin^3(2t)}\)
Multiply by \(\displaystyle \frac{8}{8}:\L\;\;\frac{8}{8}\,\cdot\frac{t^3\cdot\cos^3(2t)}{\sin^3(2t)} \;= \;\frac{1}{8}\cdot\cos^3(2t)\,\cdot\,\L\frac{8t^3}{\sin^3(2t)}\;=\;\frac{1}{8}\cdot\cos^3(2t)\,\cdot\,\L\left(\frac{2t}{\sin(2t)}\right)^3\)
Then: \(\displaystyle \L\,\lim_{2t\to0}\,\left[\frac{1}{8}\cdot\cos^3(2t)\cdot\left(\frac{2t}{\sin(2t)}\right)^3\right] \;=\;\frac{1}{8}\cdot1^3\cdot1^3\;=\;\frac{1}{8}\)