<s>[INST] How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
 [/INST] $$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(1+\frac{\ln x}{(2x+1)^2})(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)!}{(2x+1)^2(x)!} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)(x+1)}{(2x+1)^2} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{x^2+3x+1}{4x^2+4x+1} + 0 =$$ 
$$\lim\limits_{x \to +\infty} \frac{1+\frac{3}{x}+\frac{1}{x^2}}{4+\frac{4}{x}+\frac{1}{x^2}} =\frac{1}{4}$$
The only missing part is 
$$\lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = 0$$
For this you need to notice that
$$\frac{4^{x+1}}{(x+1)!} = \frac{4 \cdot 4^{x}}{(x+1)x!}=\frac{4}{x+1}\frac{4^{x}}{x!} < \frac{4}{x}\frac{4^{x}}{x!}$$
Since $\lim\limits_{n \to +\infty} \frac{4^n}{x^n}=0$ we have $\lim\limits_{x \to +\infty} \frac{4^{x+1}}{(x+1)!}=0$ as well
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