Evaluate the following limit. If you use LHospital

illusiia
2020-12-22
Answered

Evaluate the following limit. If you use LHospital

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okomgcae

Answered 2020-12-23
Author has **93** answers

Evaluate limits:

$\underset{x\to \mathrm{\infty}}{lim}(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x})$

$\underset{x\to a}{lim}[f(x)\pm g(x)]=\underset{x\to a}{lim}f(x)\pm \underset{x\to a}{lim}g(x)$

$\underset{x\to \mathrm{\infty}}{lim}(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x})=\underset{x\to \mathrm{\infty}}{lim}(3\cdot {2}^{1-x})+\underset{x\to \mathrm{\infty}}{lim}({x}^{2}\cdot {2}^{1-x})$

$=3\cdot \underset{x\to \mathrm{\infty}}{lim}({2}^{1-x})+2\cdot \underset{x\to \mathrm{\infty}}{lim}({x}^{2}\cdot {2}^{-x})$

$=3\cdot \underset{x\to \mathrm{\infty}}{lim}({2}^{1-x})+2\cdot \underset{x\to \mathrm{\infty}}{lim}({x}^{2}\frac{1}{{2}^{x}})$

Take$3\cdot \underset{x\to \mathrm{\infty}}{lim}({2}^{1-x})$

Apply exponent rule-${a}^{x}={e}^{\mathrm{ln}({a}^{x})}={e}^{x\mathrm{ln}(a)}$

$=3\cdot \underset{x\to \mathrm{\infty}}{lim}({e}^{(1-x)\mathrm{ln}(2)})$

$=3\cdot \underset{x\to \mathrm{\infty}}{lim}({2}^{1-x})$

$=3\cdot \underset{x\to \mathrm{\infty}}{lim}({e}^{x\mathrm{ln}(2)})$

Apply the Limit Chain Rule:

$g(x)=x\mathrm{ln}(2),f(u)={e}^{u}$

$=-\mathrm{\infty}\cdot \mathrm{ln}(2)$

$=-\mathrm{\infty}$

$=3\cdot lim({e}^{x\mathrm{ln}(2)})=3\cdot {e}^{-\mathrm{\infty}}$

$=3\cdot 0$

$=0$

Take$2\cdot \underset{x\to \mathrm{\infty}}{lim}({x}^{2}\frac{1}{{2}^{x}}$

if$\sum {a}_{n}$ converges, then $\underset{n\to \mathrm{\infty}}{lim}({a}_{n})=0$

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.$\underset{x\to \mathrm{\infty}}{lim}(\frac{{x}^{2}}{{2}^{x}})=0$

$=2\cdot 0$

$=0$

$\underset{x\to \mathrm{\infty}}{lim}(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x})=3\cdot \underset{x\to \mathrm{\infty}}{lim}({2}^{1-x})+2\cdot \underset{x\to \mathrm{\infty}}{lim}({x}^{2}\frac{1}{{2}^{x}})$

$=0+0$

$=0$

Result:$\underset{x\to \mathrm{\infty}}{lim}(3\cdot {2}^{1-x}+{x}^{2}\cdot {2}^{1-x})=0$

Take

Apply exponent rule-

Apply the Limit Chain Rule:

Take

if

Apply ratio test and check weather series is convergent or divergent.

If

If

diverges.

Result:

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