# [SOLVED] Trig limit Question

• Sep 26th 2009, 07:11 PM
Noxide
[SOLVED] Trig limit Question
Can someone please confirm that this expression's limit DNE. I'm wasting too much paper to try and find out.

lim x-->0 (tan3x)/(tan5x)

Also does anyone have any tips on knowing where to stop for when a trig function has a limit that doesn't exist. I find that i'm just spinning my wheels on a lot of the questions that don't have limits.

Thanks!
• Sep 26th 2009, 08:08 PM
redsoxfan325
Quote:

Originally Posted by Noxide
Can someone please confirm that this expression's limit DNE. I'm wasting too much paper to try and find out.

lim x-->0 (tan3x)/(tan5x)

Also does anyone have any tips on knowing where to stop for when a trig function has a limit that doesn't exist. I find that i'm just spinning my wheels on a lot of the questions that don't have limits.

Thanks!

It does indeed exist!

$\lim_{x\to0}\frac{\tan(3x)}{\tan(5x)}=\lim_{x\to0} \frac{\sin(3x)\cos(5x)}{\sin(5x)\cos(3x)}$ $=\left(\lim_{x\to0}\frac{5x\sin(3x)}{3x\sin(5x)}\c dot\frac{3x}{5x}\right)\left(\lim_{x\to0}\frac{\co s(5x)}{\cos(3x)}\right)$ $=\left(\lim_{x\to0}\frac{\sin(3x)}{3x}\right)\left (\lim_{x\to0}\frac{5x}{\sin(5x)}\right)\left(\lim_ {x\to0}\frac{3x}{5x}\right)\cdot 1$ $=1\cdot1\cdot\frac{3}{5}=\boxed{\frac{3}{5}}$

Alternatively, if you know L'Hopital's Rule, you could just do:

$\lim_{x\to0}\frac{\tan(3x)}{\tan(5x)}=\lim_{x\to0} \frac{3\sec^2(3x)}{5\sec^2(5x)} = \boxed{\frac{3}{5}}$
• Sep 26th 2009, 08:57 PM
Noxide
Thanks.
• Sep 26th 2009, 09:05 PM
redsoxfan325
Quote:

Originally Posted by Noxide
I don't think I quite understand what you did in the expression following the 2nd equals sign. It would be great it you, or anyone could explain this.

$\lim_{x\to0}\frac{\sin(3x)\cos(5x)}{\sin(5x)\cos(3 x)}=\lim_{x\to0}\left(\frac{3x\cdot5x\cdot\sin(3x) }{3x\cdot5x\cdot\sin(5x)}\cdot\frac{\cos(5x)}{\cos (3x)}\right)$ $=\left(\lim_{x\to0}\frac{5x\cdot\sin(3x)}{3x\cdot\ sin(5x)}\cdot\frac{3x}{5x}\right)\left(\lim_{x\to0 }\frac{\cos(5x)}{\cos(3x)}\right)$

by the Limit product rule: $\lim_{x\to a}f(x)g(x)=\left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a}g(x)\right)$ assuming both limits exist.
• Sep 26th 2009, 09:54 PM
Noxide
Yes, thanks again. The answer came to me a few seconds after I hit the post button. Another question did come up. How did you simplify

limx-->0 (cos(5x)/cos(3x))

and am I correct in saying that it became the expression limx--> 3x/5x?
If so, how did it become that?
• Sep 26th 2009, 09:58 PM
redsoxfan325
Quote:

Originally Posted by Noxide
Yes, thanks again. The answer came to me a few seconds after I hit the post button. Another question did come up. How did you simplify

limx-->0 (cos(5x)/cos(3x))

and am I correct in saying that it became the expression limx--> 3x/5x?
If so, how did it become that?

$\lim_{x\to0}\frac{\cos(5x)}{\cos(3x)}$ became $1$. You can get that just by plugging in zero. (Remember that $\cos(0)=1$.)

The $\lim_{x\to0}\frac{3x}{5x}$ came about as a result of my second usage of the limit product rule. I broke up

$\lim_{x\to0}\frac{5x\sin(3x)}{3x\sin(5x)}\cdot\fra c{3x}{5x}$

into

$\left(\lim_{x\to0}\frac{\sin(3x)}{3x}\right)\left( \lim_{x\to0}\frac{5x}{\sin(5x)}\right)\left(\lim_{ x\to0}\frac{3x}{5x}\right)$
• Sep 26th 2009, 10:03 PM
Noxide
Thank you. I really appreciate your explanations.
• Sep 26th 2009, 10:13 PM
redsoxfan325
You're welcome.