# Homework Help: Limit of a function

1. Dec 11, 2008

### tomboi03

For -1 < x < 1, x$$\neq$$0 let
f(x)= $$\frac{\sqrt{1+x}-1}{x}$$

i. Prove that
lim f(x)
x$$\rightarrow$$0
exists and find it
(There is an easy proof and you get no credit for applying "hospital's rule")
Part (i) shows that f can be continued to a continuous function on (-1,1) if we assign this limit to be f(0) (this is assumed in subseguent parts)

ii. Show that the limit
lim 1/x $$\int$$ f(t) dt
x$$\rightarrow$$0
exists

iii. Determine constants a0, a1, a2, a3 so that
$$\int$$ f(t) dt = a0 + a1x + a2x2 + a3x3 + x3 $$\rho$$(x)
(this integral goes from... 0 to x i couldn't figure out a way to put it on the integral.)
where
lim $$\rho$$(x) =0
x$$\rightarrow$$0

Thanks

2. Dec 11, 2008

### gabbagabbahey

Start with part (i). What must be true of $$\lim_{x\to0^{+}} f(x)$$ and $$\lim_{x\to0^{-}} f(x)$$ for $$\lim_{x\to0} f(x)$$ to exist?

3. Dec 11, 2008

### tomboi03

if both the integral approaches the same constant from the negative and positive side, then the function will approach that constant.

but, where do i go from there?

4. Dec 11, 2008

### gabbagabbahey

Integral; what integral? Don't you mean 'limit'?

For the limit to exist, both one-sided limits must approach the same value.

That means that if both one-sided limits approach the same value, then the limit exists.

So.....do both one-sided limits approach the same value?

Last edited: Dec 11, 2008
5. Dec 11, 2008

### HallsofIvy

For (i) you might want to rationalize the numerator.

For (ii) remember that an integral is always differentiable. What is the "difference quotient" for this?

For (iii) I don't understand what is wanted if f itself is not given explicitely.

6. Dec 11, 2008

### Pere Callahan

I assume f means the function of (i) (continued continuously at zero)

7. Dec 11, 2008

### HallsofIvy

Oh, yes, I see now. i, ii, and iii all come after the definition of f. But no integration is necessary for ii.

8. Dec 11, 2008

### HallsofIvy

tomboi03, think about a Taylor's series expansion.