# Math Help - limit help

1. ## limit help

I got stuck on this limit
lim as x approaches -1 ln(2+x)/(x+1)
I got the limit down to lim as x approaches -1 ((1/2+x))/(x+1) by using L'Hospitals rule, but got stuck from there. I am having trouble getting an answer of 1. (answer given in book thanks.)

2. You're getting there.

$\lim_{x\rightarrow{-1}}\frac{ln(2+x)}{x+1}$

L'Hopital: $\lim_{x\rightarrow{-1}}\frac{1}{x+2}$

Now, see how it's 1?.

3. $D_x\ln(x+2)=\frac1{x+2}$ & $(x+1)'=1$

Is that what you need?

4. Originally Posted by davecs77
I got stuck on this limit
lim as x approaches -1 ln(2+x)/(x+1)
We can avoid the use of L'Hopital's rule.

First,
$\lim_{x\to -1} \frac{\ln (x+2)}{x+1} = \lim_{x\to 0} \frac{\ln (x+1)}{x}$

Notice the derivative of $f(x) = \ln (x+1)$ at zero is $f'(0) = 1$.

But, using the definition of derivative we get,
$\lim_{x\to 0} \frac{\ln (x+1) - \ln (0+1)}{x- 0} = \lim_{x\to 0}\frac{\ln (x+1)}{x} =f'(0)= 1$

5. Originally Posted by ThePerfectHacker
We can avoid the use of L'Hopital's rule.

First,
$\lim_{x\to -1} \frac{\ln (x+2)}{x+1} = \lim_{x\to 0} \frac{\ln (x+1)}{x}$

Notice the derivative of $f(x) = \ln (x+1)$ at zero is $f'(0) = \frac{1}{2}$.

But, using the definition of derivative we get,
$\lim_{x\to 0} \frac{\ln (x+1) - \ln (0+1)}{x- 0} = \lim_{x\to 0}\frac{\ln (x+1)}{x} =f'(0)= \frac{1}{2}" alt="\lim_{x\to 0} \frac{\ln (x+1) - \ln (0+1)}{x- 0} = \lim_{x\to 0}\frac{\ln (x+1)}{x} =f'(0)= \frac{1}{2}" />
$\lim_{x\to 0} \frac{\ln (x+1) - \ln (0+1)}{x- 0} = \lim_{x\to 0}\frac{\ln (x+1)}{x} =f'(1)= 1" alt="\lim_{x\to 0} \frac{\ln (x+1) - \ln (0+1)}{x- 0} = \lim_{x\to 0}\frac{\ln (x+1)}{x} =f'(1)= 1" />
as $f^{\prime}(1) = 1$. This gives us back the correct answer for the limit.