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!

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.
