# Homework Help: Limit. Does it exist or not?

1. May 30, 2013

### mathgeek69

1. lim as x->o+ ( (sqrt(x+sinx))(lnx))

3. The attempt at a solution:

lim as x->o+ ( (sqrt(x+sinx))(lnx))
= lim as x->o+ (sqrt(x+sinx)) x lim as x->o+(lnx)
= 0(-∞)
= 0

Last edited by a moderator: May 6, 2017
2. May 30, 2013

### haruspex

0 × ∞ is indeterminate. When you get that result in a limits question it tells you you need to use a subtler approach.
What does sin(x) approximate as x tends to zero? Does that allow you to simplify it a little?
Can you see a variable substitution that would then allow you to lose the sqrt?

3. May 30, 2013

### mathgeek69

as sin(x) approaches 0 as x approaches 0.

That didn't help me as I still resolve to 0*(-∞) but as you say we need a different approach.

I did try to do" let y = (x+sin(x))^2 " and try to take that approach but I didnt know what to do after as now I am introducing y all of a sudden and dont know how ln x is affected.

4. May 30, 2013

### haruspex

No, I was asking for some simpler function that it approximates on its way to 0.

5. May 30, 2013

### mathgeek69

It approximates cos(x+ (pi/2)) ?

6. May 30, 2013

### haruspex

No, I said simpler. Here's a clue, do you know what the limit of sin(x)/x is as x tends to 0?

7. May 30, 2013

### mathgeek69

0/0 ?

8. May 31, 2013

### HallsofIvy

Then you need to review (or learn) limits.

9. May 31, 2013

### haruspex

0/0 is useless for the same reason that 0 × ∞ is useless. In these situation you get nowhere by taking the limit of each part independently.
What is the tangent to sin(x) at x = 0?