# Evaluate limits involving indeterminate forms

• August 21st 2009, 06:32 AM
dalbir4444
Evaluate limits involving indeterminate forms
Evaluate the limit:

lim t->0 (t^9)/(tan 2t)^9

Do I have to keep applying l'hopital's rule 9 times or is there another way to do it?
• August 21st 2009, 06:46 AM
ynj
no, you may use Taylor.
$\lim\frac{t^9}{\tan^9 2t}=\lim\frac{t^9}{(2t+o(t^2))^9}=\lim\frac{t^9}{5 12t^9+o(t^9)}=lim\frac{1}{512+\frac{o(t^9)}{t^9}}= \frac{1}{512}$
• August 21st 2009, 08:05 AM
Soroban
Helolo, dalbir4444!

Use the theorem: . $\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1 \qquad \lim_{\theta\to0}\frac{\theta}{\sin\theta} \:=\:1$

Quote:

Evaluate: . $\lim_{t\to0}\frac{t^9}{(\tan 2t)^9}$
$\text{We have: }\;\frac{t^9}{\dfrac{(\sin2t)^9}{(\cos2t)^9}} \;=\; \frac{t^9(\cos2t)^9}{(\sin2t)^9}$

Multiply by $\frac{2^9}{2^9}\!:\quad \frac{2^9}{2^9}\cdot \frac{t^9(\cos2t)^0}{(\sin2t)^9} \;=\;\frac{(\cos2t)^9}{2^9}\cdot\frac{2^9t^9}{(\si n2t)^9} \;=\;\left(\frac{\cos2t}{2}\right)^9 \cdot\frac{(2t)^9}{(\sin2t)^9}$ . $= \;\left(\frac{\cos2t}{2}\right)^9\left(\frac{2t}{\ sin2t}\right)^9$

$\text{Take the limit: }\;\lim_{t\to0}\left(\frac{\cos2t}{2}\right)^9\lef t(\frac{2t}{\sin2t}\right)^9 \;=\; \lim_{t\to0}\underbrace{\left(\frac{\cos2t}{2}\rig ht)}_{\text{this is }\frac{1}{2}}\,\!\!^9\cdot\lim_{t\to0}\underbrace{ \left(\frac{2t}{\sin2t}\right)}_{\text{This is 1}}\,\!\!^9$

. . . . . . . . . . $= \;\left(\frac{1}{2}\right)^9\cdot1^9 \;=\;\frac{1}{512}$