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Limit with L'Hopital's rule

105
3
1. Homework Statement

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

2. Homework 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,.
 

Dick

Science Advisor
Homework Helper
26,258
618
1. Homework Statement

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

2. Homework 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,.
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?
 
105
3
$$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$
$$\lim_{x\rightarrow+\infty}(ln \frac{2x}{x-4})$$

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

Science Advisor
Homework Helper
26,258
618
$$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$
$$\lim_{x\rightarrow+\infty}(ln \frac{2x}{x-4})$$

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

Staff Emeritus
Science Advisor
Homework Helper
12,794
1,665
$$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$
$$\lim_{x\rightarrow+\infty}(ln \frac{2x}{x-4})$$

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?
 
105
3
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.
 
105
3
Is there another way to do it?
 

ehild

Homework Helper
15,360
1,766
Is there another way to do it?
[tex]\lim_{x\rightarrow \infty} \frac{2x}{x-4}=\lim_{x\rightarrow \infty}\frac{2}{1-4/x}[/tex]
4/x goes to zero ...
 
105
3
[tex]\lim_{x\rightarrow \infty} \frac{2x}{x-4}=\lim_{x\rightarrow \infty}\frac{2}{1-4/x}[/tex]
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

Homework Helper
15,360
1,766
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 :)
 

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