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 2 years ago
would lim(x approches infinity) (sqrt(x^2+1))/(2x+1) be DNE since it would basically be infinity/infinity?
 2 years ago
would lim(x approches infinity) (sqrt(x^2+1))/(2x+1) be DNE since it would basically be infinity/infinity?

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redham
 2 years ago
Best ResponseYou've already chosen the best response.0That is indeterminate form, so you can't say it doesn't exist. You will have to manipulate it to use L'hopital's

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.3\[\lim_{x\rightarrow\infty}\frac{\sqrt{x^2+1}}{2x+1}=\lim_{x\rightarrow\infty}\frac{\sqrt{\frac {x^2}{x^2}+\frac1{x^2}}}{\frac{2x}x+\frac1x}=\lim_{x\rightarrow\infty}\frac{\sqrt{1}}{2}=\frac12\]

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.3@redham L'hopital's is not the best idea.

Valpey
 2 years ago
Best ResponseYou've already chosen the best response.0No, you can map a limit here.\[\lim_{x \rightarrow \infty}\Big(\frac{\sqrt{x^2+1}}{2x+1}\Big)=\frac{\lim_{x \rightarrow \infty}\Big(\frac{d\sqrt{x^2+1}}{dx}\Big)}{\lim_{x \rightarrow \infty}\Big(\frac{d(2x+1)}{dx}\Big)}\]

DRico
 2 years ago
Best ResponseYou've already chosen the best response.0how do i simplify to sqrt(1)/2

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.3\[\lim_{x\rightarrow\infty}\frac{\sqrt{\frac {x^2}{x^2}+\frac1{x^2}}}{\frac{2x}x+\frac1x}=\lim_{x\rightarrow\infty}\frac{\sqrt{1+1/x^2}}{2+1/x}\] \(1/x^2\) and \(1/x\) are almost 0 when x goes to infinity. so only 1in numerator and 2 in denominator are left.

redham
 2 years ago
Best ResponseYou've already chosen the best response.0@klimenkov I was actually thinking of squaring the denominator under the radical and then using L'Hopital's. Not from the start

DRico
 2 years ago
Best ResponseYou've already chosen the best response.0one more if y'all are still here how to calculate the derivative of f(x)=sinx

DRico
 2 years ago
Best ResponseYou've already chosen the best response.0i know the derivative of sinx is cosx but i dont know how to prove it with the derivative funtion

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.3\[\lim_{\Delta x\rightarrow0}\frac{\sin(x+\Delta x)\sin x}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{\sin x \cos\Delta x+\cos x\sin \Delta x\sin x}{\Delta x}=\]\[\lim_{\Delta x\rightarrow0}\frac{\sin x (\cos\Delta x1)}{\Delta x}+\lim_{\Delta x\rightarrow0}\frac{\cos x\sin \Delta x}{\Delta x}=\cos x\] Try to get it. Why is it so? Why does the first limit equals to 0 and why does the second equals to cos x?
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