# Homework Help: Basic Limit problems

1. Dec 6, 2007

### thrive

1. lim sin(1/x) No clue what to do here
x>0

2. lim xsin(1/x) i believe the answer is infinity
x>00

3. lim (tan(5x)/sec(5x))*((cos(3x)/4x))
x>0

2. Dec 6, 2007

### Office_Shredder

Staff Emeritus
Just to start you off with the first one, try doing the substitution v=1/x

3. Dec 6, 2007

### thrive

undefined

The crux of the whole problem is that I don't know how to handle Sin(undefined)

edit: answer is DNE, im stupid :< as sin function approaches 0, there is no value that it locks onto...it just keeps oscillating up and down like the sin function does, infinitely.

Last edited: Dec 6, 2007
4. Dec 6, 2007

### Office_Shredder

Staff Emeritus
For the second one then, try writing it as sin(1/x)/(1/x), then do the substitution v=1/x.

Note that when you do a substitution like this, it's not completely rigorous as 1/x can go to +/- infinity as v goes to zero, and similiar problems going in reverse, but it gives a very strong intuitive feel for what's going on, and with care can be made rigorous

5. Dec 6, 2007

### thrive

wow im completely lost with that explanation. All i got was that sin(0)/0 would occur...aka indeterminate form

6. Dec 6, 2007

### kuahji

If all else fails, use numerical methods to find the limit.

7. Dec 6, 2007

### Dick

That is so right.

8. Dec 7, 2007

### thrive

if all else fails go to physics forums

9. Dec 7, 2007

### Office_Shredder

Staff Emeritus
You should either know the limit of sin(x)/x as x goes to zero, or you should look it up, because it's quite useful

10. Dec 7, 2007

### dynamicsolo

In following OfficeShredder's suggestion, you got lim u->0 [sin(u)/u]. You are correct is that the Limit Laws would give an indeterminate form. However, there is a proof concerning this particular limit; have you had that in your book or course?

You will need the result for lim u->0 [sin(u)/u] in order to do this one. To make your work somewhat easier, rewrite this in terms of just sine and cosine functions. You will have one place where you need the (sin u)/u limit , given that u can be a function of x, but you will have to make an adjustment by multiplication and division of constants.

11. Dec 7, 2007

### thrive

we already learned that lim x->0 [sin(x)/x] = 1

i just do not understand how the lim x->0 of [xsin(1/x)] relates to this

12. Dec 7, 2007

### Office_Shredder

Staff Emeritus
Hold on, above you had the limit as x went to infinity. Assuming that's what it was, then xsin(1/x)=sin(1/x)/(1/x)=sin(v)/v where v is going to zero.

If x is actually going to 0, then |xsin(1/x)|<=|x| for all x, so you can use the squeeze theorem to find the limit

13. Dec 8, 2007

### thrive

I understand this, however how can v just change from going to infinity to go to 0?

14. Dec 8, 2007

### Dick

If x->infinity, v=1/x->0.

15. Dec 8, 2007

### thrive

so as x approaches infinity, v=1 as x approaches 0? I still don't understand this reasoning...

16. Dec 8, 2007

### Dick

No, no. v=(1/x) -> 0. If v is defined as 1/x, then as the variable x approaches infinity, the variable v approaches zero. By changing variables from x to v, you change an 'approaches infinity' type limit to an 'approaches zero' type limit.

17. Dec 8, 2007

### thrive

ok, so when showing my work I just go from:

step 1: lim X-> 00 [xsin(1/x)]
step 2: lim X-> 0 [sin(1/x)/(1/x)]
step 3: = 1

so once we change the form from multiplying by x to dividing by 1/x we also change the limit going to infinity to go to 0?

18. Dec 8, 2007

### Dick

Yes, that's it. It's a little confusing to have the two different x's. Makes it clearer to say the v=1/x substitution changes the limit to sin(v)/v with v->0.

19. Dec 8, 2007

### Office_Shredder

Staff Emeritus