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_{xrightarrowinfty}(3cdot2^{1-x}+x^2cdot2^{1-x})

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

2020-12-23
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: $$\lim_{x\rightarrow\infty}(3\cdot2^{1-x}+x^2\cdot2^{1-x})=0$$