Prove that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$

Question

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.

Answer 1

Divide numerator and denominator by $2^n$. It's not hard to show that $\frac{x^2}{2^n}\to 0$ (use L'Hôpital's rule, for example) to see $$\lim_{x\to\infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}=\lim_{x\to\infty}\frac{2+(x+1)^2/2^x}{1+x^2/2^x}=2$$

Answer 2

$$\lim_{x\to \infty}\frac{2\cdot (2^{x}+x^2)+(x+1)^2-2x^2}{2^x+x^2}$$ $$=2+\lim_{x\to \infty}\frac{(x+1)^2-2x^2}{2^x+x^2}$$

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

Answer 3

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: $$=1+\lim_{x\to \infty}\frac{2^x\ln 2+2}{2^x\ln 2+2x}$$ $$=1+\lim_{x\to \infty}\frac{2^x(\ln 2)^2}{2^x(\ln 2)^2+2}$$ $$=1+\lim_{x\to \infty}\frac{2^x(\ln 2)^3}{2^x(\ln 2)^3}$$ $$=1+\lim_{x\to \infty}1=1+1=2$$