# Thread: A Couple Of Limits

1. ## A Couple Of Limits

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.

2. $\displaystyle 2)\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {x + 1} \right)}} {x}\underbrace = _{LH}\mathop {\lim }\limits_{x \to 0} \frac{1} {{x + 1}} = 1$

3. 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

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

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

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

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

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