[SOLVED] Trig limit Question

**Noxide** · September 26th 2009, 06:11 PM

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!

**redsoxfan325** · September 26th 2009, 07:08 PM

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:

$\lim_{x\to0}\frac{\tan(3x)}{\tan(5x)}=\lim_{x\to0} \frac{3\sec^2(3x)}{5\sec^2(5x)} = \boxed{\frac{3}{5}}$

**Noxide** · September 26th 2009, 07:57 PM

Thanks.

**redsoxfan325** · September 26th 2009, 08:05 PM

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.

**Noxide** · September 26th 2009, 08:54 PM

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?

**redsoxfan325** · September 26th 2009, 08:58 PM

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

$\left(\lim_{x\to0}\frac{\sin(3x)}{3x}\right)\left( \lim_{x\to0}\frac{5x}{\sin(5x)}\right)\left(\lim_{ x\to0}\frac{3x}{5x}\right)$

**Noxide** · September 26th 2009, 09:03 PM

Thank you. I really appreciate your explanations.

**redsoxfan325** · September 26th 2009, 09:13 PM

You're welcome.

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