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.

$$\lim_{x\to \infty}\frac{2\cdot (2^{x}+x^2)+(x+1)^2-2x^2}{2^x+x^2}$$
$$=2+\lim_{x\to \infty}\frac{(x+1)^2-2x^2}{2^x+x^2}$$
($\frac{\infty}{\infty}$)form so using L-Hospital's rule twice:
$$=2+\lim_{x\to \infty}\frac{-2}{2^x (ln2)^2+2}$$
$$=2$$
