Proving Limit of Function for -1 < x < 1, x≠0: Prove & Find Lim

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For -1 < x < 1, x\neq0 let
f(x)= \frac{\sqrt{1+x}-1}{x}

i. Prove that
lim f(x)
x\rightarrow0
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\rightarrow0
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\rightarrow0


I don't know how to go about this at all.

Please help me out
Thanks
 
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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?
 
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?
 
tomboi03 said:
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?

Integral; what integral? :confused: 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?
 
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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.
 
HallsofIvy said:
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.

I assume f means the function of (i) (continued continuously at zero)
 
Oh, yes, I see now. i, ii, and iii all come after the definition of f. But no integration is necessary for ii.
 
tomboi03, think about a Taylor's series expansion.
 
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