Question

Limit of tan5x/sin2x as x approaches 0.

This is what I have so far, but I'm not sure what to do next. Any ideas?

lim (x-->0) tan(5x) / sin(2x)
= lim (x-->0) [sin(5x)/cos(5x)] / sin(2x)

Answer 1

Okay, so far so good. Recall the special limit(s),
\[\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1\]
for \(a\not=0\). To use this fact, you have to rewrite your trig expression so that you can split up the limit of a product into a product of limits.

Answer 2

So would I rewrite it as:

lim (x-->0) tan(5x) / sin(2x)
= lim (x-->0) [sin(5x)/cos(5x)] / sin(2x)
= lim (x-->0) [sin(5x)/sin(2x)] * lim (x-->0) [1/cos(5x)]

?

Answer 3

Yes, that's another intermediate step. How would you go about introducing the proper \(ax\) terms?

Answer 4

I'm not quite sure where to go from there, since there's a 'sin' in both the numerator and denominator. I assume I have to get rid of the 'cos' portion somehow?

Answer 5

The cosine limit can be evaluated by substituting directly.
\[\lim_{x\to0}\frac{1}{\cos5x}=\frac{1}{\cos0}=1\]
so it disappears.

As for the sine limits, currently you have
\[\lim_{x\to0}\frac{\sin ax}{\sin bx}\]
To get it into the form we want (so that we can use that handy limit), you want to introduce factors of \(ax\) and \(bx\) that don't fundamentally alter the expressions here. One way to do that is to multiply by 1, which appears in the forms \(\dfrac{\text{something}}{\text{something}}\).
\[\lim_{x\to0}\frac{\sin ax}{\sin bx}\cdot\frac{ax}{ax}\cdot\frac{bx}{bx}\]

Answer 6

I'm still kind of confused about the multiplying bit. Do I multiple both the num. & den. by 5x? And if so, I don't really understand how it gets me closer to the special limit form.

Sorry! Just having a hard time understanding this since I've never really done it before.

Answer 7

We want the special limit form,
\[\lim_{x\to0}\frac{\sin ax}{ax}=1\]
(or its reciprocal).

What we have:
\[\lim_{x\to0}\frac{\sin ax}{\sin bx}\cdot\frac{ax}{ax}\cdot\frac{bx}{bx}\]
What we should do:
\[\lim_{x\to0}\frac{\sin ax}{ax}\cdot\frac{bx}{\sin bx}\cdot\frac{ax}{bx}\]
This is still the same expression. Now we can split up limit of a product as a product of limits:
\[\left(\lim_{x\to0}\frac{\sin ax}{ax}\right)\cdot\left(\lim_{x\to0}\frac{bx}{\sin bx}\right)\cdot\left(\lim_{x\to0}\frac{ax}{bx}\right)\]

Answer 8

Ohh, I see. So then I'd get 1x1x(5x/2x) = 5/2. Is that correct?

Answer 9

Right, that's all there is to it!

Answer 10

Awesome, thank you so much for your help!

Answer 11

You're welcome!