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anonymous
 4 years ago
what is the limit of sqrt (x^210x+1)x as x approaches infinity
anonymous
 4 years ago
what is the limit of sqrt (x^210x+1)x as x approaches infinity

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\rightarrow +\infty}\sqrt{x^210x+1}x=\]\[=\lim_{x\rightarrow +\infty}\left(\sqrt{x^210x+1}x\right)\cdot\frac{\sqrt{x^210x+1}+x}{\sqrt{x^210x+1}+x}=\]\[=\lim_{x\rightarrow +\infty}\frac{x^210x+1x^2}{\sqrt{x^210x+1}+x}=\]\[=\lim_{x\rightarrow +\infty}\frac{10x+1}{x\sqrt{110/x+1/x^2}+x}=\lim_{x\rightarrow +\infty}\frac{10x+1}{2x}=5\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let f(x)=x^3−2x. Calculate the difference quotient f(2+h)−f(2)/h for h=.1 h=.01 h=−.01 h=−.1 if you can demonstrate how to do one, I will learn to do the rest

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[h=0.1\]\[f(2)=4\quad,\quad f(2+h)=f(2+0.1)=f(2.1)=5.061\]\[\frac{f(2+h)f(2)}{h}=\frac{f(2.1)f(2)}{0.1}=\frac{1.061}{0.1}=10.61\]
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