# Evaluate the following limit. If you use L'Hospital Rule, be sure to indicate when you are using it, and why it applies. lim_{xrightarrowinfty}(3cdot2^{1-x}+x^2cdot2^{1-x})

Evaluate the following limit. If you use LHospital
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Evaluate limits:
$\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)$
$\underset{x\to a}{lim}\left[f\left(x\right)±g\left(x\right)\right]=\underset{x\to a}{lim}f\left(x\right)±\underset{x\to a}{lim}g\left(x\right)$
$\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)=\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}\right)+\underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\cdot {2}^{1-x}\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)+2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\cdot {2}^{-x}\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)+2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\frac{1}{{2}^{x}}\right)$
Take $3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)$
Apply exponent rule- ${a}^{x}={e}^{\mathrm{ln}\left({a}^{x}\right)}={e}^{x\mathrm{ln}\left(a\right)}$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({e}^{\left(1-x\right)\mathrm{ln}\left(2\right)}\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({e}^{x\mathrm{ln}\left(2\right)}\right)$
Apply the Limit Chain Rule:
$g\left(x\right)=x\mathrm{ln}\left(2\right),f\left(u\right)={e}^{u}$
$=-\mathrm{\infty }\cdot \mathrm{ln}\left(2\right)$
$=-\mathrm{\infty }$
$=3\cdot lim\left({e}^{x\mathrm{ln}\left(2\right)}\right)=3\cdot {e}^{-\mathrm{\infty }}$
$=3\cdot 0$
$=0$
Take $2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\frac{1}{{2}^{x}}$
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 or divergent.
If $|\frac{{a}_{n+1}}{{a}_{n}}|\le q$ eventually for some 0<q<1, then $\sum _{n=1}^{\mathrm{\infty }}|{a}_{n}|$ converges.
If $|\frac{{a}_{n+1}}{{a}_{n}}|>1$ eventually then $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$
diverges. $\underset{x\to \mathrm{\infty }}{lim}\left(\frac{{x}^{2}}{{2}^{x}}\right)=0$
$=2\cdot 0$
$=0$
$\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)+2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\frac{1}{{2}^{x}}\right)$
$=0+0$
$=0$
Result: $\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)=0$