# Limits problem

1. Aug 16, 2010

### Misr

1. The problem statement, all variables and given/known data
[PLAIN]http://img517.imageshack.us/img517/2516/62548919.jpg [Broken]

I'm hopeless to solve this problem
you can give me a hint before solving the problem!

Last edited by a moderator: May 4, 2017
2. Aug 16, 2010

### Staff: Mentor

You might try rewriting this as
$$\lim_{n \to \infty}\frac{(A + (1/n))^4 - A^4}{1/n}$$

In this form, the expression has the indeterminate form [0/0], so you can use L'Hopital's Rule on it.

3. Aug 16, 2010

### Misr

still can't solve it ...we can factorize the numerator but still makin no sense....Can u make it simpler?

Last edited: Aug 16, 2010
4. Aug 16, 2010

### iamalexalright

Did you try L'Hopital's rule?

5. Aug 16, 2010

### Staff: Mentor

Do you know how to use L'Hopital's Rule?

A slightly different approach is this, with h = 1/n
$$\lim_{h \to 0^+}\frac{(A + h)^4 - A^4}{h}$$

6. Aug 16, 2010

### Misr

No...I don't know anything about this rule we don't study it at school and they can't give us a problem on a rule which we haven't studied yet :(

7. Aug 16, 2010

### iamalexalright

Well, if you use Mark44's substitution you wont need to use L'Hopital's rule

8. Aug 16, 2010

### Hurkyl

Staff Emeritus
You need to show us what work you have done on the problem, Misr, even if it's just thoughts about it. It is unacceptable to just post a question and say "I can't do it".

And a reminder for homework helpers: our goal is to help posters do a problem, not to give them answers. The OP has gotten plenty of advice, and his own idea that he didn't post the work for sounds like a very workable approach too. If he doesn't start showing that he's done anything with any of that, then it is inappropriate to give further "help".

9. Aug 16, 2010

### Misr

10. Aug 17, 2010

### hunt_mat

Have you tried expanding:
$$(A+n^{-1})^{4}=A^{4}+4\frac{A^{3}}{n}+6\frac{A^{2}}{n^{2}}+4\frac{A}{n^{3}}+\frac{1}{n^4}$$
I think that this will solve your problem.