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:

*

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


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


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


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


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


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


*$e^{\infty} = \infty$
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!
 A: 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.
A: 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.
$$\to e^{-\infty}=0$$
A: 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.
A: Note that $$\lim_{u\to \infty}\left( 1-\frac{1}{u}\right)^{u}=\frac{1}{e}$$
A: 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 $0$.
