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

[MarkFL]

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.

 

[MarkFL]

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:

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