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

AI Thread Summary
The limit as x approaches infinity for the expression ((3x-1)/(3x-4))^2x can be solved without L'Hôpital's rule by rewriting the fraction as 1 + (3/(3x-4)). This leads to the limit being expressed as (1 + (3/u))^(2/3)(u + 4) after substitution. By applying properties of limits and exponents, the limit can be separated into two parts, one of which evaluates to 1, and the other simplifies to (e^3)^(2/3). Ultimately, the limit is determined to be e^2.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
Hello Miras Tussupov,

We are given to evaluate:

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

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

$$\frac{3x-1}{3x-4}=\frac{3x-4+3}{3x-4}=1+\frac{3}{3x-4}$$

And now we have:

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

Next, let's use the substitution:

$$x=3x-4$$

to obtain:

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

Applying the properties of exponents, we may write:

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

Applying the properties of limits, we may write:

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

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

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

Using the well-known formula:

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

we now have:

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

Hence, we may conclude:

$$\lim_{x\to\infty}\left(\left(\frac{3x-1}{3x-4} \right)^{2x} \right)=e^2$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top