A Couple Of Limits

**trigoon** · December 5th 2008, 07:39 AM

Im having trouble solving the following limits, any help would be appreciated

Ive tried and for the second one I kept on getting that it DNE (Does not exist) but I've been told the answer is 1.

**Nacho** · December 5th 2008, 08:12 AM

$<br /> 2)\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {x + 1} \right)}}<br /> {x}\underbrace = _{LH}\mathop {\lim }\limits_{x \to 0} \frac{1}<br /> {{x + 1}} = 1<br />$

**trigoon** · December 5th 2008, 08:24 AM

Thanks! It seems I was doing the right thing but made a mistake when taking the derivative of ln(x+1) no idea how I made such a silly mistake

**galactus** · December 5th 2008, 09:32 AM

Here's a way to look at the second one if you like.

Use the series for $\frac{ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\frac{x^{4}}{5}-\frac{x^{5}}{6}+......$

$\lim_{x\to 0}\left[1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\frac{x^{4}}{5}-\frac{x^{5}}{6}+......\right]$

See now where the 1 comes in?. As all the x terms drop due to the 0, all that's left is 1.

**trigoon** · December 5th 2008, 09:42 AM

Thanks

I got #2 now, its just #1 and #3 thats bugging me.

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