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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!

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  • $\begingroup$ Think of this: $\frac{3x+2}{4x+3} < 1$ for $x>0$...then you take Infinitiv power...this will lead to zero $\endgroup$ Dec 17 '20 at 16:19
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    $\begingroup$ $lim_{x \to \infty} \frac{4x +3}{x+1} =4 \ne 1$. So from 5$^{\text{th}}$ to 6$^{\text{th}}$ step is wrong. $\endgroup$ Dec 17 '20 at 16:21
  • $\begingroup$ 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 $\displaystyle\lim_{x\to\infty}\frac{3x + 2}{4x + 3} < 1$. $\endgroup$ Dec 17 '20 at 16:25
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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.

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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.

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

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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.

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

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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 $0$.

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