anonymous
  • anonymous
would lim(x approches infinity) (sqrt(x^2+1))/(2x+1) be DNE since it would basically be infinity/infinity?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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klimenkov
  • klimenkov
What is DNE?
anonymous
  • anonymous
does not exist
anonymous
  • anonymous
That is indeterminate form, so you can't say it doesn't exist. You will have to manipulate it to use L'hopital's

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More answers

klimenkov
  • klimenkov
\[\lim_{x\rightarrow\infty}\frac{\sqrt{x^2+1}}{2x+1}=\lim_{x\rightarrow\infty}\frac{\sqrt{\frac {x^2}{x^2}+\frac1{x^2}}}{\frac{2x}x+\frac1x}=\lim_{x\rightarrow\infty}\frac{\sqrt{1}}{2}=\frac12\]
klimenkov
  • klimenkov
@redham L'hopital's is not the best idea.
Valpey
  • Valpey
No, you can map a limit here.\[\lim_{x \rightarrow \infty}\Big(\frac{\sqrt{x^2+1}}{2x+1}\Big)=\frac{\lim_{x \rightarrow \infty}\Big(\frac{d\sqrt{x^2+1}}{dx}\Big)}{\lim_{x \rightarrow \infty}\Big(\frac{d(2x+1)}{dx}\Big)}\]
anonymous
  • anonymous
how do i simplify to sqrt(1)/2
klimenkov
  • klimenkov
\[\lim_{x\rightarrow\infty}\frac{\sqrt{\frac {x^2}{x^2}+\frac1{x^2}}}{\frac{2x}x+\frac1x}=\lim_{x\rightarrow\infty}\frac{\sqrt{1+1/x^2}}{2+1/x}\] \(1/x^2\) and \(1/x\) are almost 0 when x goes to infinity. so only 1in numerator and 2 in denominator are left.
anonymous
  • anonymous
@klimenkov I was actually thinking of squaring the denominator under the radical and then using L'Hopital's. Not from the start
anonymous
  • anonymous
thnaks to all of you
anonymous
  • anonymous
one more if y'all are still here how to calculate the derivative of f(x)=sinx
anonymous
  • anonymous
i know the derivative of sinx is cosx but i dont know how to prove it with the derivative funtion
klimenkov
  • klimenkov
\[\lim_{\Delta x\rightarrow0}\frac{\sin(x+\Delta x)-\sin x}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{\sin x \cos\Delta x+\cos x\sin \Delta x-\sin x}{\Delta x}=\]\[\lim_{\Delta x\rightarrow0}\frac{\sin x (\cos\Delta x-1)}{\Delta x}+\lim_{\Delta x\rightarrow0}\frac{\cos x\sin \Delta x}{\Delta x}=\cos x\] 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?
anonymous
  • anonymous
im not exactly sure...
Valpey
  • Valpey
http://www.zweigmedia.com/RealWorld/trig/triglim.html

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