# Limit with L'Hopital's rule

Maged Saeed

## Homework Statement

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

## 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,.

Homework Helper

## Homework Statement

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

## 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,.

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?

Maged Saeed
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
:)

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

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
:)

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.

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

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
:)

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

Maged Saeed
Why do you want to use l'Hopital's rule inside the log?

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

Maged Saeed
Is there another way to do it?

Mentor
Is there another way to do it?
See Dick's post #4.

Homework Helper
Is there another way to do it?
$$\lim_{x\rightarrow \infty} \frac{2x}{x-4}=\lim_{x\rightarrow \infty}\frac{2}{1-4/x}$$
4/x goes to zero ...

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

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).

Homework Helper
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).

Yes :)

Maged Saeed