# Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$

Limit i want to solve: $$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$$

This is how i started solving this limit:

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

2. $$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$$

3. $$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$$

4. $$\left(1- \frac {x+1}{4x+3}\right)^x$$

5. $$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$$

6. $$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$$

7. $$e^{\infty} = \infty$$

answer i got is $$\infty$$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong? Thanks!

• Think of this: $\frac{3x+2}{4x+3} < 1$ for $x>0$...then you take Infinitiv power...this will lead to zero Commented Dec 17, 2020 at 16:19
• $lim_{x \to \infty} \frac{4x +3}{x+1} =4 \ne 1$. So from 5$^{\text{th}}$ to 6$^{\text{th}}$ step is wrong. Commented Dec 17, 2020 at 16:21
• Not exactly right, as $\displaystyle\lim_{x\to\infty}\bigg(1 - \frac{1}{x}\bigg)^{x} = \frac{1}{e}$ even though $1 -\frac{1}{x} < 1$ for $x>0$. A better way to think about it would be $\displaystyle\lim_{x\to\infty}\frac{3x + 2}{4x + 3} < 1$. Commented Dec 17, 2020 at 16:25

Now @Infinity_hunter has explained your error, note that $$\frac34-\frac{3x+2}{4x+3}=\frac{1}{4(4x+3)}>0$$, so $$0<\left(\frac{3x+2}{4x+3}\right)^x<\left(\frac34\right)^x$$ proves the limit is $$0$$ by squeezing.

Being $$\lim_{x \to \infty} \left(\frac{3 x + 2}{4 x + 3}\right)^{x} = \lim_{x \to \infty} e^{\ln{\left(\left(\frac{3 x + 2}{4 x + 3}\right)^{x} \right)}}= e^{\lim_{x \to \infty} x \ln{\left(\frac{3 x + 2}{4 x + 3} \right)}}=e^{\lim_{x \to \infty} x \color{blue}{\ln{\left(\lim_{x \to \infty} \frac{3 x + 2}{4 x + 3} \right)}}}$$

If you multiply and divide by $$x$$ the ratio $$\frac{3 x + 2}{4 x + 3}=\frac{\frac{3 x + 2}x}{\frac{4 x + 3}x}=\frac{3+\frac{2}x}{4+\frac{3}x}$$ you have

$$\lim_{x \to \infty} \frac{3+\frac{2}x}{4+\frac{3}x}=\frac 34$$

But if we remember that $$\ln \frac 34<0$$, the limit for $$x\to \infty$$ is $$-\infty$$ (to the exponent of the Napier's number) i.e.

$$\to e^{-\infty}=0$$

We have:

$$\lim_{x\to\infty}\bigg(\frac{3x + 2}{4x + 3}\bigg)^{x} = \lim_{x\to\infty}\bigg(\frac{3 + \frac{2}{x}}{4 + \frac{3}{x}}\bigg)^{x} = \bigg(\frac{3}{4}\bigg)^{\infty} = \boxed{0}$$

Your mistake was in assuming that $$\frac{4x+3}{x + 1}\to\infty$$ as $$x\to\infty$$, which meant you could apply the definition of $$e$$. However, $$\frac{4x + 3}{x+1}\to 4$$. Remember that a rational function only diverges if the degree of the numerator is greater than the degree of the denominator.

Note that $$\lim_{u\to \infty}\left( 1-\frac{1}{u}\right)^{u}=\frac{1}{e}$$

First of all let investigate terms of inside the parentheses. When $$x$$ approaches to infinity $$\displaystyle \frac{3x+2}{4x+3}=\frac{3}{4}$$ then the limit transformed into $$\displaystyle \lim_{x \to \infty} \left(\frac{3}{4}\right)^{x}$$. Rewritre $$\frac{3^x}{4^x}$$. Obviously denomiter is greater than numerator. So $$\lim$$ approaches to $$0$$.