# Thread: limit problem

1. ## limit problem

2) Evaluate: $\displaystyle lim_{x \to 0} (\frac{1}{ln(x+1)} - \frac{1}{x})$
I don't know how to do this. Please show me a step-by-step solution

2. The only solution I can think of right now involves Maclaurin series. Have you learnt that yet?

$\displaystyle \frac{1}{\ln(x+1)}-\frac{1}{x}=\frac{x-\ln(x+1)}{x\ln(x+1)}$

Then replace $\displaystyle \ln(x+1)$ with its Maclaurin series:

$\displaystyle \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O(x^5)$

3. Replacing ln(x+1) with the McLaurin series, you wll still have a problem with both numerator and denominator going to 0.

So use L'Hopital's rule.
$\displaystyle \frac{d(x-ln(x+1))}{dx}= 1- \frac{1}{x+1}$
$\displaystyle \frac{d(x ln(x+1))}{dx}= ln(x+1)+ \frac{x}{x+1}$

Now the limit, as x goes to 0, of each of those, is 0 so do it again:
$\displaystyle \frac{d(1- \frac{1}{x+1})}{dx}= \frac{1}{(x+1)^2}$
$\displaystyle \frac{d(ln(x+1)+ \frac{x}{x+1})}{dx}= \frac{1}{x+1}+ \frac{1}{x+1}- \frac{x}{(x+1)^2}$

The limit, as x goes to 0, of the numerator is 1 while the limit of the denominator is 2. The original limit is 1/2.

4. Originally Posted by HallsofIvy
Replacing ln(x+1) with the McLaurin series, you wll still have a problem with both numerator and denominator going to 0.
You can factor out $\displaystyle x^2$ to get something like $\displaystyle \frac{\frac{1}{2}+O(x)}{1+O(x)}$

5. Okay, good point.