# Limits with trig functions

1. Feb 4, 2004

### burge

I am having trouble with the following problems:

1) lim as x -> 0 of (sin 3x)/2x

2) lim as x -> 0 of (tan 5x)/(sin 2x)

3) lim as x -> 0 of (sin²3x)/2x

4) lim as h -> 0 of [(h+x)³ -x³]/h

5) lim as h -> 0 of [1/(x+h) - 1/x]/h

Last edited: Feb 4, 2004
2. Feb 4, 2004

### himanshu121

These are basic formulas

for eg u can write sinx~x when x->0 similarly for tanx

What Have u covered so far in Limits this will help us to give u better explanations

3. Feb 4, 2004

### burge

We know that sin and cos are continuous,
the lim as x-> 0 (sin (x))/x = 1
the lim as x-> 0 (1 - cos(x))/x = 0
and some basic trig identities

4. Feb 4, 2004

### HallsofIvy

It may be that your problem is tjat "(sin 3x)/2x" has different coefficients for the x inside and outside of the sine. That no big deal. First take out the "1/2": (1/2)(sin3x)/x and the multiply both numerator and denominator by 3: (3/2)(sin3x)/3x Think of the "3x" as u and the problem is (3/2) (sin u)/u. Since u= 3x goes to 0 when x does, what is the limit of that as u goes to 0?

5. Feb 4, 2004

### burge

I got that one now, I'm really stuck on (sin² 3x)/2x
Also, I know there is a trick for finding it in the calculator, since there is no sin² button, but I don't remember it.

6. Feb 4, 2004

### himanshu121

write it as sin3x*sin3x/2x and now find the limit it would be zero

7. Feb 4, 2004

### matt grime

1) lim as x -> 0 of (sin 3x)/2x

set u = 3x as halls of ivy suggests

2) lim as x -> 0 of (tan 5x)/(sin 2x)

tan(5x) = sin(5x)/cos(5x) and it suffices to find out what sin(5x)/sin(2x) tends to so you could work it all out in terms of sin2x

or look at the end of the post

3) lim as x -> 0 of (sin²3x)/2x

4) lim as h -> 0 of [(h+x)³ -x³]/h

expand the bracket

5) lim as h -> 0 of [1/(x+h) - 1/x]/h

have you done basic algebra? cos that again is 'just do the manipulation' - think before asking, you'll learn a lot more

in general you want to use l'Hopital's rule.