For -1 < x < 1, x[tex]\neq[/tex]0 let(adsbygoogle = window.adsbygoogle || []).push({});

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 a_{0}, a_{1}, a_{2}, a_{3}so that

[tex]\int[/tex] f(t) dt = a_{0}+ a_{1}x + a_{2}x^{2}+ a_{3}x^{3}+ x^{3}[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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# Homework Help: Limit of a function

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