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Discussion Response Post to: Limits

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23 Mar 2008 12:36:33 IST

Subject: Re:Limits

hsbhatt (5799)

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nice work hegde avare

You can do it without taking log.

Again setting $x = \frac{1} {n}.$

$\frac {f(a+x)} {f(a)} = 1 + \frac{f(a+x) - f(a)} {f(a)} = 1+x\frac{f$ in the limit $x \rightarrow 0$

Hence the limit becomes

$\lim_{x\rightarrow 0} (1+x\frac{f$

Time wounds all heels

this reply: 15 points (with 3 Olaaa!! Perrrfect answer.

in 3 votes ) [?]

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