# Cancellation Limit

1. Jul 25, 2010

### themadhatter1

1. The problem statement, all variables and given/known data
Find the limit.

$$\lim_{x\rightarrow0}\frac{\frac{1}{x+1}-1}{x}$$

2. Relevant equations

3. The attempt at a solution

I have to do this analytically. Although, I know that the limit is supposed to be -1 from a graphing approach. When you substitute in 0 for x you get 0/0. How do I get this function into a determinate form? I cant factor and I'm not sure what to multiply the top and bottom by.

I tried $$\frac{\frac{1}{x+1}+1}{\frac{1}{x+1}+1}$$

and I got

$$\frac{1-(x^2+2x+4)}{x(x+1)(x+2)}$$ which is still in indeterminate form.

2. Jul 25, 2010

### eumyang

I wouldn't do that. Try multiplying the original fraction by
$$\frac{x+1}{x+1}$$
instead. Something should eventually cancel.

69

3. Jul 25, 2010

### Mute

Do you know, or can you use, L'Hopital's rule?

In this case, L'Hopital's rule would take the form

If
$$\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{0}{0}$$
but
$$\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L$$
where L is a finite number,
then
$$\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L$$

4. Jul 25, 2010

### Bohrok

You can get a common denominator with the terms in the numerator and simplify, then you should get a fraction that's not indeterminate anymore.

5. Jul 25, 2010

### themadhatter1

No I dont, I'm just starting calculus, but that looks interesting. I'm sure I'll learn it later.

Thanks, that works perfectly. I know where I went wrong.

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