# Evaluate the following limits. lim_{xrightarrowinfty}(3cdot2^{1-x}+x^2cdot2^{1-x})

Limits and continuity
Evaluate the following limits. $$\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})$$

2021-02-10

Evaluate limit:
$$\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})$$
$$\lim_{x\rightarrow a}[f(x)+-g(x)]=\lim_{x\rightarrow a}f(x)+-\lim_{x\rightarrow a}g(x)$$
$$\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})=\lim_{x\rightarrow\infty}(3\cdot2^{1-x})+\lim_{x\rightarrow\infty}(x^2\cdot2^{1-x})$$
$$=3\cdot\lim_{x\rightarrow\infty}(2^{1-x})+2\cdot\lim_{x\rightarrow\infty}(x^2\cdot2^{-x})$$
$$=3\cdot\lim_{x\rightarrow\infty}(2^{1-x})+2\cdot\lim_{x\rightarrow\infty}(x^2\cdot\frac{1}{2^x})$$
Take $$3\cdot\lim_{x\rightarrow\infty} (2^{1-x})$$
Apply exponent rule $$a^x=e^{\ln a^x}=e^{x\cdot\ln a}$$
$$=3\cdot\lim_{x\rightarrow\infty}(e^{(1-x)\cdot\ln2})$$
$$=3\cdot\lim_{x\rightarrow\infty}2^{1-x}$$
$$=3\cdot\lim_{x\rightarrow\infty}e^{x\cdot\ln2}$$
Apple the Limit Chain Rule:
$$g(x)=x\ln(2),f(u)=e^u$$
$$=-\infty\cdot\ln2$$
$$=-\infty$$
$$3\cdot\lim_{x\rightarrow\infty}e^{x\cdot\ln2}=3\cdot e^{-\infty}$$
$$=3\cdot0$$
$$=0$$
Take $$2\cdot\lim_{x\rightarrow\infty}x^2\cdot\frac{1}{2^x}$$
if sum $$a_n$$ converges,then $$\lim_{n\rightarrow\infty}(a_n)=0$$
Apply ratio test and check weather series is convergent of divergent
if $$|\frac{a_{n+1}}{a_n}|\leq q$$ eventually for some 0, then $$\sum_{n=1}^x|a_n|$$ converges,
if $$|\frac{a_{n+1}}{a_n}|>1$$ eventually then $$\sum_{n=1}^x a_n$$ diverges
$$\lim_{x\rightarrow\infty}(\frac{x^2}{2^x})=0$$
$$=2\cdot0$$
$$=0$$
$$\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})=3\cdot\lim_{x\rightarrow\infty}(2^{1-x})+2\cdot\lim_{x\rightarrow\infty}(x^2\cdot\frac{1}{2^x})$$
$$=0+0$$
$$=0$$
Result:
$$3\cdot\lim_{x\rightarrow\infty}(2^{1-x})+2\cdot\lim_{x\rightarrow\infty}(x^2\cdot\frac{1}{2^x})=0$$