Solve Limit w/o L'Hôpital: Miras Tussupov's Q on Yaoo Answers

In summary, we are asked to find the limit of ((3x-1)/(3x-4))^2x as x approaches infinity without using L'Hopital's rule. We can rewrite the expression and apply properties of limits and exponents to obtain the final answer of e^2.
  • #1
MarkFL
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Here is the question:

Solve limit when x approaches infinity.?

How can I solve limit when x approaches infinity without using L'Hopital's rule?

lim(((3x-1)/(3x-4))^2x)

I have posted a link there to this thread so the OP can view my work.
 
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  • #2
Hello Miras Tussupov,

We are given to evaluate:

\(\displaystyle L=\lim_{x\to\infty}\left(\left(\frac{3x-1}{3x-4} \right)^{2x} \right)\)

First, let's rewrite the innermost expression as follows:

\(\displaystyle \frac{3x-1}{3x-4}=\frac{3x-4+3}{3x-4}=1+\frac{3}{3x-4}\)

And now we have:

\(\displaystyle L=\lim_{x\to\infty}\left(\left(1+\frac{3}{3x-4} \right)^{2x} \right)\)

Next, let's use the substitution:

\(\displaystyle x=3x-4\)

to obtain:

\(\displaystyle L=\lim_{u\to\infty}\left(\left(1+\frac{3}{u} \right)^{\frac{2}{3}(u+4)} \right)\)

Applying the properties of exponents, we may write:

\(\displaystyle L=\lim_{u\to\infty}\left(\left(1+\frac{3}{u} \right)^{\frac{8}{3}}\cdot\left(1+\frac{3}{u} \right)^{\frac{2}{3}u} \right)\)

Applying the properties of limits, we may write:

\(\displaystyle L=\lim_{u\to\infty}\left(\left(1+\frac{3}{u} \right)^{\frac{8}{3}} \right)\cdot\lim_{u\to\infty}\left(\left(1+\frac{3}{u} \right)^{\frac{2}{3}u} \right)\)

The first limit is a determinate for and is equal to 1, and the second limit may be rewritten as:

\(\displaystyle L=\left(\lim_{u\to\infty}\left(\left(1+\frac{3}{u} \right)^{u} \right) \right)^{\frac{2}{3}}\)

Using the well-known formula:

\(\displaystyle \lim_{x\to\infty}\left(\left(1+\frac{k}{x} \right)^x \right)=e^k\)

we now have:

\(\displaystyle L=\left(e^3 \right)^{\frac{2}{3}}=e^2\)

Hence, we may conclude:

\(\displaystyle \lim_{x\to\infty}\left(\left(\frac{3x-1}{3x-4} \right)^{2x} \right)=e^2\)
 

FAQ: Solve Limit w/o L'Hôpital: Miras Tussupov's Q on Yaoo Answers

What is "Solve Limit w/o L'Hôpital: Miras Tussupov's Q on Yahoo Answers"?

"Solve Limit w/o L'Hôpital: Miras Tussupov's Q on Yahoo Answers" is a specific math problem posted on the popular question and answer website, Yahoo Answers. It involves solving a limit without using the L'Hôpital's rule, a commonly used technique in calculus.

Why is this problem important?

This problem challenges individuals to think critically and come up with alternative methods to solve a limit without relying on a common rule. It helps improve problem-solving skills and promotes creativity in mathematical thinking.

What are some common strategies to solve this problem?

Some common strategies to solve this problem include using Taylor series, substitution, and algebraic manipulation. Other approaches may involve using geometric interpretations or considering special cases.

Can this problem be solved using L'Hôpital's rule?

No, the whole point of this problem is to find alternative methods to solve a limit without using L'Hôpital's rule. However, it is possible to solve this problem using L'Hôpital's rule, but it defeats the purpose of the challenge.

Where can I find the solution to this problem?

The solution to this problem can be found on Yahoo Answers, where it was originally posted. However, it is always recommended to try solving the problem on your own before looking for the solution.

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