# Can't find limit

1. Sep 15, 2009

### honestrosewater

Problem
Find the limit.

$$\lim_{x \rightarrow 0} \frac{\frac{1}{x + 1} - 1}{x}$$

My attempt
I'm totally stuck. domain(f) = R - {0}. Setting g: R - {-1} --> R, g(x) = 1/(x+1) - 1 and h: R --> R, h(x) = x, g and h are both continuous, but g(0) = 0 and h(0) = 0. It looks like f(0) is just a hole. Perhaps I will try factoring again. My book says the limit is -1, but I don't see how it expects me to find it.

2. Sep 15, 2009

### tiny-tim

Hi honestrosewater!

erm

what's 1/(x+1) - 1 ?

( alternatively, use l'Hôpital's rule )

3. Sep 15, 2009

### honestrosewater

1/(x + 1) - 1 = -x/(x+1)? Does that help me somehow? Sorry, I am sure it's something simple, but I cannot see it. Oh... right.

$$\left(\frac{-x}{x + 1}\right)\left(\frac{1}{x}\right) = \frac{-1}{x + 1} = g(x)$$

g is continuous and x != 0 implies g(x) = f(x), so lim(x --> 0) f(0) = lim(x --> 0) g(0) = -1. Hah.

Thanks for this tip. I will remember it later. Unfortunately, passing my test means applying the algorithms that the book has taught us, and we have not covered derivatives yet or been taught that rule. I have already been warned about using theorems that I am not supposed to know.

Thanks! :^)