# Limit of Arcsin Approaching Infinity

1. Feb 28, 2008

### tsw303

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

limit approaching infinity: (arcsin(x))/(x)

= 0

Question is: Why? The 'Sandwich Theorem' 0=[(arcsinx)/x]=0 gives this
solution, but looking at the graph of (arcsinx)/x , this appears
impossible.

2. Relevant equations

lim x->OO [arcsin(x)] - {DNE)
lim x->OO [(arcsinx)/x] - {DNE/OO}

lim x->OO [-Pi/2x] = lim x->OO [(arcsinx)/x] = lim x->OO [Pi/2x]
=> 0 = [(arcsinx)/x] = 0

=> lim x->OO [(arcsinx)/x] = 0

HOWEVER: (-Pi/2)<[arcsinx]<(Pi/2) ... SO ...
... (x) never reaches (OO) for (Pi/2x) to reach the limit x->OO (Pi/OO)
= 0.

3. The attempt at a solution

The sandwich theorem gives and answer of "0". Maple 11 gives the same answer, so does my teacher. This just doesn't seem possible.

2. Feb 28, 2008

### foxjwill

The only way you can get "0" is if you're using complex limits. I'm pretty sure. Let me check up on that.

Last edited: Feb 28, 2008
3. Feb 29, 2008

### chaoseverlasting

If you use, L'Hospitals Rule, you get a $$\frac{1}{\sqrt{1-x^2}}$$, for x tending to infinity. Thats a zero, but a complex one, dont know if its even applicable here...

4. Feb 29, 2008

### d_leet

arcsin(x) isn't defined for x outside of [-1, 1] considering only the reals. So unless x is complex, I don't think this limit exists.

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