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

Evaluate the following limits. $\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)$
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Derrick

Evaluate limit:
$\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)$
$\underset{x\to a}{lim}\left[f\left(x\right)+-g\left(x\right)\right]=\underset{x\to a}{lim}f\left(x\right)+-\underset{x\to a}{lim}g\left(x\right)$
$\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)=\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}\right)+\underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\cdot {2}^{1-x}\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)+2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\cdot {2}^{-x}\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)+2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\cdot \frac{1}{{2}^{x}}\right)$
Take $3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)$
Apply exponent rule ${a}^{x}={e}^{\mathrm{ln}{a}^{x}}={e}^{x\cdot \mathrm{ln}a}$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({e}^{\left(1-x\right)\cdot \mathrm{ln}2}\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}{2}^{1-x}$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}{e}^{x\cdot \mathrm{ln}2}$
Apple the Limit Chain Rule:
$g\left(x\right)=x\mathrm{ln}\left(2\right),f\left(u\right)={e}^{u}$
$=-\mathrm{\infty }\cdot \mathrm{ln}2$
$=-\mathrm{\infty }$
$3\cdot \underset{x\to \mathrm{\infty }}{lim}{e}^{x\cdot \mathrm{ln}2}=3\cdot {e}^{-\mathrm{\infty }}$
$=3\cdot 0$
$=0$
Take $2\cdot \underset{x\to \mathrm{\infty }}{lim}{x}^{2}\cdot \frac{1}{{2}^{x}}$
if sum ${a}_{n}$ converges,then $\underset{n\to \mathrm{\infty }}{lim}\left({a}_{n}\right)=0$
Apply ratio test and check weather series is convergent of divergent
if $|\frac{{a}_{n+1}}{{a}_{n}}|\le q$ 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
$\underset{x\to \mathrm{\infty }}{lim}\left(\frac{{x}^{2}}{{2}^{x}}\right)=0$
$=2\cdot 0$
$=0$
$\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)+2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\cdot \frac{1}{{2}^{x}}\right)$
$=0+0$
$=0$
Result:
$3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)+2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\cdot \frac{1}{{2}^{x}}\right)=0$