<s>[INST] 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!
 [/INST] Note that $$\lim_{u\to \infty}\left( 1-\frac{1}{u}\right)^{u}=\frac{1}{e}$$
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