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

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Homework Help Overview

The discussion revolves around proving the limit of the function f(x) = (√(1+x) - 1) / x as x approaches 0, specifically for the interval -1 < x < 1, x ≠ 0. Participants are tasked with showing the existence of this limit and exploring related integrals and constants.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conditions under which the limit exists, particularly focusing on one-sided limits. There are suggestions to rationalize the numerator for part (i) and considerations about the differentiability of integrals for part (ii). Some participants express confusion about the requirements for part (iii) without explicit definitions.

Discussion Status

The discussion is active, with participants offering guidance on rationalizing expressions and considering Taylor series expansions. There is an ongoing exploration of the implications of one-sided limits and the nature of the integral involved in part (ii). Multiple interpretations of the problem are being examined, particularly regarding the continuity of the function and the requirements for the limits.

Contextual Notes

Participants note that the original function f must be defined continuously at zero for subsequent parts of the problem. There is also mention of homework constraints that discourage the use of certain methods, such as L'Hôpital's rule.

tomboi03
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For -1 < x < 1, x[tex]\neq[/tex]0 let
f(x)= [tex]\frac{\sqrt{1+x}-1}{x}[/tex]

i. Prove that
lim f(x)
x[tex]\rightarrow[/tex]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 [tex]\int[/tex] f(t) dt
x[tex]\rightarrow[/tex]0
exists

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


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 [tex]\lim_{x\to0^{+}} f(x)[/tex] and [tex]\lim_{x\to0^{-}} f(x)[/tex] for [tex]\lim_{x\to0} f(x)[/tex] 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?
 
Last edited:
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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