Problem with limit

**atlantis84** · September 5th 2009, 06:54 PM

Any help to the following?

$\lim_{x \rightarrow 0} \left(\frac{ln (x+1) }{x} \right)^{\frac{2}{X}}$

**mr fantastic** · September 5th 2009, 09:42 PM

Originally Posted by atlantis84

Any help to the following?

$\lim_{x \rightarrow 0} \left(\frac{ln (x+1) }{x} \right)^{\frac{2}{X}}$

One possible approach:
1. Substitute the following Maclaurin series: $\ln (x + 1) = x - \frac{x^2}{2} + \frac{x^3}{3} - ....$ .

2. Simplify the result.

3. Consider $\lim_{x \rightarrow 0} e^{\ln \left( 1 - \frac{x}{2} + \frac{x^2}{3} - .... \right)^{2/x}} = \lim_{x \rightarrow 0} e^{2 \ln \left( 1 - \frac{x}{2} + \frac{x^2}{3} - .... \right)/x}$ .

4. Consider the limit of the exponent using l'Hopital's rule.

Alternatively, there may be a clever substitution that can first be made.

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