Evaluate the following limits. If you use l'Hospital's Rule, be sure to indicate

Evaluate the following limits. If you use lHospitals Rule, be sure to indicate when yyou are using it, and why it applies.
a) $\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)$
b) $\underset{x\to {0}^{+}}{lim}{\left(1+5x\right)}^{\frac{2}{x}}$
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Walker Funk
Evaluate limits:
$\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)$
$\underset{x\to a}{lim}\left[f\left(x\right)±g\left(x\right)\right]=\underset{x\to a}{lim}f\left(x\right)±\underset{x\to a}{lim}g\left(x\right)$
$\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x}\right)=\underset{x\to \mathrm{\infty }}{lim}\left(3\cdot {2}^{1-x}\right)+\underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\cdot {2}^{-x}\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)+2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\frac{1}{{2}^{x}}\right)$
Take $3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)$
Apply exponent rule ${a}^{x}={e}^{\mathrm{ln}\left({a}^{x}\right)}={e}^{x\mathrm{ln}\left(a\right)}$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({e}^{\left(1-x\right)\mathrm{ln}\left(2\right)}\right)\right)$
$=3\cdot \underset{x\to \mathrm{\infty }}{lim}\left({2}^{1-x}\right)$
$=3\cdot \underset{x\to -\mathrm{\infty }}{lim}\left({e}^{x\mathrm{ln}\left(2\right)}\right)$
Apply the Limit Chain Rule:
$g\left(x\right)=x\mathrm{ln}\left(2\right),f\left(u\right)={e}^{u}$
$=-\mathrm{\infty }\cdot \mathrm{ln}\left(2\right)$
$=-\mathrm{\infty }$
$3\cdot \underset{x\to -\mathrm{\infty }}{lim}\left({e}^{x\mathrm{ln}\left(2\right)}\right)=3\cdot {e}^{-\mathrm{\infty }}$
$=3\cdot 0$
$=0$
Take $2\cdot \underset{x\to \mathrm{\infty }}{lim}\left({x}^{2}\frac{1}{{2}^{x}}\right)$
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, the
Not exactly what you’re looking for?
Lauren Fuller
b) Evaluate limits:
$\underset{x\to {0}^{+}}{lim}{\left(1+5x\right)}^{\frac{2}{x}}$
Apply exponent rule:
${a}^{x}={e}^{\mathrm{ln}\left({a}^{x}\right)}={e}^{x\cdot \mathrm{ln}\left(a\right)}$
${\left(1+5x\right)}^{\frac{2}{x}}={e}^{\frac{2}{x}\mathrm{ln}\left(1+5x\right)}$
Apply the limit Chain Rule:
$g\left(x\right)=\frac{2}{x}\mathrm{ln}\left(1+5x\right),f\left(u\right)={e}^{u}$
$\underset{x\to {0}^{+}}{lim}g\left(x\right)=\underset{x\to {0}^{+}}{lim}\frac{2}{x}\mathrm{ln}\left(1+5x\right)$
$=2\cdot \underset{x\to {0}^{+}}{lim}\left(\frac{\mathrm{ln}\left(1+5x\right)}{x}\right)$
$\underset{x\to {0}^{+}}{lim}\left(\frac{\mathrm{ln}\left(1+5x\right)}{x}\right)=\frac{0}{0}$
LHopital