L'Hopital's Rule

**nanotek887** · December 2nd 2006, 09:17 AM

Could someone help me use L'Hopital's Rule to find the following limit?

lim (as x approaches 0) (1+(3/x))^x
Thank you.

**TD!** · December 2nd 2006, 09:30 AM

You need to get an indeterminate form such as 0/0 or inf/inf to apply l'Hôpital.

$\left( {1 + \frac{3}{x}} \right)^x = e^{\ln \left( {1 + \frac{3}{x}} \right)^x } = e^{x\ln \left( {1 + \frac{3}{x}} \right)}$

Now we can use the fact that:

$<br /> \mathop {\lim }\limits_{x \to 0} e^{x\ln \left( {1 + \frac{3}{x}} \right)} = e^{\mathop {\lim }\limits_{x \to 0} x\ln \left( {1 + \frac{3}{x}} \right)} <br />$

Now you have two possibilities, one is:

$<br /> x\ln \left( {1 + \frac{3}{x}} \right) = \frac{{\ln \left( {1 + \frac{3}{x}} \right)}}{{\frac{1}{x}}}<br />$

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