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

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

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

klimenkovBest ResponseYou've already chosen the best response.3
@redham L'hopital's is not the best idea.
 one year ago

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

DRicoBest ResponseYou've already chosen the best response.0
how do i simplify to sqrt(1)/2
 one year ago

klimenkovBest 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.
 one year ago

redhamBest 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
 one year ago

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

DRicoBest ResponseYou've already chosen the best response.0
i know the derivative of sinx is cosx but i dont know how to prove it with the derivative funtion
 one year ago

klimenkovBest 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?
 one year ago

ValpeyBest ResponseYou've already chosen the best response.0
http://www.zweigmedia.com/RealWorld/trig/triglim.html
 one year ago
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