Prove that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$ Could someone please show me the proof that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$
I have no idea where to begin with this one.
Thanks.

Notice, we have $$\lim_{x\to \infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}$$
$$=\lim_{x\to \infty}\frac{2\cdot 2^{x}+x^2+2x+1}{2^x+x^2}$$
$$=\lim_{x\to \infty}\frac{(2^{x}+x^2)+(2^x+2x+1)}{2^x+x^2}$$
$$=1+\lim_{x\to \infty}\frac{2^x+2x+1}{2^x+x^2}$$
Using L-Hospital's rule 3 times:
$$=1+\lim_{x\to \infty}\frac{2^x\ln 2+2}{2^x\ln 2+2x}$$
$$=1+\lim_{x\to \infty}\frac{2^x(\ln 2)^2}{2^x(\ln 2)^2+2}$$
$$=1+\lim_{x\to \infty}\frac{2^x(\ln 2)^3}{2^x(\ln 2)^3}$$
$$=1+\lim_{x\to \infty}1=1+1=2$$
