# Thread: [SOLVED] Trig limit Question

1. ## [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!

2. 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:

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

3. Thanks.

4. 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.

5. 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?

6. 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

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

7. Thank you. I really appreciate your explanations.

8. You're welcome.

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# limx-0 (tan(3x)/tan(5x))

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