# Limit with L'Hopital's rule

1. Dec 20, 2014

### Maged Saeed

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

How can I fount the following limit using L'H'opetal rule?
$$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$

2. Relevant equations

3. The attempt at a solution
I tried to use the low
$$\lim_{x\rightarrow a} f(x)=\lim_{x\rightarrow a}e^{ln(f(x))}=e^l$$
But it seems to be useless

Many thanks for Help,.

2. Dec 20, 2014

### Dick

Start by using some rules of logs, like ln(a)-ln(b)=ln(a/b). Can you find the limit of the a/b part?

3. Dec 20, 2014

### Maged Saeed

$$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$
$$\lim_{x\rightarrow+\infty}(ln \frac{2x}{x-4})$$

Yes I can use L'H'opetal rule to find (a/b)

$$\lim_{x\rightarrow+\infty}ln(\frac{2x}{x-4})$$
$$\lim_{x\rightarrow+\infty}ln(\frac{2}{1})$$

Oh,, it will be ln(2) .

But how can I use L'H'opetal rule inside the (ln)?

Thanks_For_Help
:)

4. Dec 20, 2014

### Dick

Because ln is a continuous function. If $\lim_{x\rightarrow+\infty} f(x)=L$ and $L>0$ then $\lim_{x\rightarrow+\infty} ln(f(x))=ln(L)$. Use l'Hopital to find the limit of the expression inside the ln.

5. Dec 20, 2014

### SteamKing

Staff Emeritus
Why do you want to use l'Hopital's rule inside the log?

6. Dec 20, 2014

### Maged Saeed

I'm supposed to solve it by L'Hopital's rule .
But I didn't think about another idea to solve it.

7. Dec 20, 2014

### Maged Saeed

Is there another way to do it?

8. Dec 20, 2014

### Staff: Mentor

See Dick's post #4.

9. Dec 21, 2014

### ehild

$$\lim_{x\rightarrow \infty} \frac{2x}{x-4}=\lim_{x\rightarrow \infty}\frac{2}{1-4/x}$$
4/x goes to zero ...

10. Dec 21, 2014

### Maged Saeed

The right answer is ln (2)

Thus , I should evaluate the limit inside the (ln) then the final answer should be put in the (ln).

11. Dec 21, 2014

Yes :)