Question

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

Answer 1

Evaluate limits:
\(\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})\)
\(\lim_{x\rightarrow a}[f(x)\pm g(x)]=\lim_{x\rightarrow a}f(x)\pm\lim_{x\rightarrow a}g(x)\)
\(\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})=\lim_{x\rightarrow\infty}(3\cdot2^{1-x})+\lim_{x\rightarrow\infty}(x^2\cdot2^{1-x})\)
\(=3\cdot\lim_{x\rightarrow\infty}(2^{1-x})+2\cdot\lim_{x\rightarrow\infty}(x^2\cdot2^{-x})\)
\(=3\cdot\lim_{x\rightarrow\infty}(2^{1-x})+2\cdot\lim_{x\rightarrow\infty}(x^2\frac{1}{2^x})\)
Take \(3\cdot\lim_{x\rightarrow\infty}(2^{1-x})\)
Apply exponent rule- \(a^x=e^{\ln(a^x)}=e^{x\ln(a)}\)
\(=3\cdot\lim_{x\rightarrow\infty}(e^{(1-x)\ln(2)})\)
\(=3\cdot\lim_{x\rightarrow\infty}(2^{1-x})\)
\(=3\cdot\lim_{x\rightarrow\infty}(e^{x\ln(2)})\)
Apply the Limit Chain Rule:
\(g(x)=x\ln(2),f(u)=e^u\)
\(=-\infty\cdot\ln(2)\)
\(=-\infty\)
\(=3\cdot\lim(e^{x\ln(2)})=3\cdot e^{-\infty}\)
\(=3\cdot0\)
\(=0\)
Take \(2\cdot\lim_{x\rightarrow\infty}(x^2\frac{1}{2^x}\)
if \(\sum a_n\) converges, then \(\lim_{n\rightarrow\infty}(a_n)=0\)
Apply ratio test and check weather series is convergent or divergent.
If \(|\frac{a_{n+1}}{a_n}|\leq q\) eventually for some 0\(\sum_{n=1}^\infty|a_n|\) converges.
If \(|\frac{a_{n+1}}{a_n}|>1\) eventually then \(\sum_{n=1}^\infty a_n\)
diverges. \(\lim_{x\rightarrow\infty}(\frac{x^2}{2^x})=0\)
\(=2\cdot0\)
\(=0\)
\(\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})=3\cdot\lim_{x\rightarrow\infty}(2^{1-x})+2\cdot\lim_{x\rightarrow\infty}(x^2\frac{1}{2^x})\)
\(=0+0\)
\(=0\)
Result: \(\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})=0\)