Using L'Hopital's Rule to Evaluate Limits

[Maged Saeed]

Homework Statement

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

Homework Equations

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,.

 

[Dick]

Homework Statement

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

Homework Equations

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]

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

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
:)

 

[Dick]

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

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.

 

[SteamKing]

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

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?

 

[Mark44]

Is there another way to do it?

See Dick's post #4.

 

[ehild]

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).

 

[ehild]

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 :)