Using L'Hopital's Rule to Evaluate Limits

In summary, to find the limit of ln(2x)-ln(x-4) as x approaches positive infinity, we can use the rule ln(a)-ln(b)=ln(a/b) and then apply l'Hopital's rule to evaluate the limit inside the ln function. The final answer is ln(2).
  • #1
Maged Saeed
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3

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,.
 
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  • #2
Maged Saeed said:

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?
 
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  • #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
:)
 
  • #4
Maged Saeed said:
$$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$
$$\lim_{x\rightarrow+\infty}(ln \frac{2x}{x-4})$$

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.
 
  • #5
Maged Saeed said:
$$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$
$$\lim_{x\rightarrow+\infty}(ln \frac{2x}{x-4})$$

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?
 
  • #6
SteamKing said:
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.
 
  • #7
Is there another way to do it?
 
  • #8
Maged Saeed said:
Is there another way to do it?
See Dick's post #4.
 
  • #9
Maged Saeed said:
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 ...
 
  • #10
ehild said:
[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).
 
  • #11
Maged Saeed said:
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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What is L'Hopital's rule?

L'Hopital's rule is a mathematical tool used to evaluate limits of indeterminate forms, where the limit of the numerator and denominator both approach 0 or infinity. It states that for a function f(x) divided by a function g(x), if the limit of f(x)/g(x) as x approaches a certain value is an indeterminate form (0/0 or infinity/infinity), then the limit of f'(x)/g'(x) is equal to the same limit of f(x)/g(x).

When should L'Hopital's rule be used?

L'Hopital's rule should only be used when the limit of a function takes on an indeterminate form. It should not be used for limits that can be evaluated using algebraic simplification or other methods.

What are the conditions for using L'Hopital's rule?

The conditions for using L'Hopital's rule are that the limit must be of the form 0/0 or infinity/infinity, and the functions in the numerator and denominator must be differentiable in a neighborhood of the limit point.

Can L'Hopital's rule be used for limits at infinity?

Yes, L'Hopital's rule can also be used for limits at infinity, as long as the limit takes on an indeterminate form.

Are there any limitations to L'Hopital's rule?

Yes, L'Hopital's rule is not applicable to all limits. It cannot be used for limits that are not indeterminate forms, such as limits that approach a finite value or limits that approach 0 or infinity at different rates. It also cannot be used for limits of oscillating functions.

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