Question

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

Limits and continuity
ANSWERED
asked 2021-02-09
Evaluate the following limits. \(\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})\)

Answers (1)

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:
\(3\cdot\lim_{x\rightarrow\infty}(2^{1-x})+2\cdot\lim_{x\rightarrow\infty}(x^2\cdot\frac{1}{2^x})=0\)

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