# Limit of a Sin Function

1. Nov 5, 2011

### Jimbo57

1. The problem statement, all variables and given/known data

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP120619i0ea2h15bci31300005d9a3422292h43h4?MSPStoreType=image/gif&s=43&w=174&h=39 [Broken]

2. Relevant equations

3. The attempt at a solution

So Wolfram says to use L'Hopital as the first step, we haven't learned anything about this yet so there has to be another way using calculus that a first year math student would know.

I would first expand the denominator of x^2-4 to (x+2)(x-2) and cancel out the x+2 in both numerator and denominator. And I'm left with:
((x-1)sin1)/(x-2)
This is where I get lost...

Any help would be much appreciated!
Jim

Last edited by a moderator: May 5, 2017
2. Nov 5, 2011

### grzz

One cannot seperate (x+2) from sin any more than one can seperate ... a bone from the mouth of a very hungry dog!

Either one 'cancels' the whole sin(x+2) or nothing at all.

3. Nov 5, 2011

### LCKurtz

You don't cancel the (x+2)'s. You should know

$$\lim_{x\rightarrow 0}\frac{\sin x}{x}$$

Use that somehow.

4. Nov 5, 2011

### grzz

Did you mention in class that lim(a x b) = lim(a) x lim(b)?

5. Nov 5, 2011

### Jimbo57

Yes, we mentioned all the above, and this was much easier after LCKurtz's suggestion:

=lim sin((x+2)/(x+2)) * lim (x-1)/(x-2)
= 1 * -3/-4
=3/4
I really need to learn how to use that equation generator thing...

Much appreciated folks! :)