Find Limit: Cancellation Homework Statement

[themadhatter1]

Homework Statement

Find the limit.

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

Homework Equations

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 can't 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.

 

[eumyang]

Homework Statement

Find the limit.

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

Homework Equations

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 can't 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}

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


69

 

[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

 

[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.

 

[themadhatter1]

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

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

I wouldn't do that. Try multiplying the original fraction by
\frac{x+1}{x+1}
instead. Something should eventually cancel.69

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