Question

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

Answer 1

What is DNE?

Answer 2

does not exist

Answer 3

That is indeterminate form, so you can't say it doesn't exist. You will have to manipulate it to use L'hopital's

Answer 4

\[\lim_{x\rightarrow\infty}\frac{\sqrt{x^2+1}}{2x+1}=\lim_{x\rightarrow\infty}\frac{\sqrt{\frac {x^2}{x^2}+\frac1{x^2}}}{\frac{2x}x+\frac1x}=\lim_{x\rightarrow\infty}\frac{\sqrt{1}}{2}=\frac12\]

Answer 5

@redham
L'hopital's is not the best idea.

Answer 6

No, you can map a limit here.\[\lim_{x \rightarrow \infty}\Big(\frac{\sqrt{x^2+1}}{2x+1}\Big)=\frac{\lim_{x \rightarrow \infty}\Big(\frac{d\sqrt{x^2+1}}{dx}\Big)}{\lim_{x \rightarrow \infty}\Big(\frac{d(2x+1)}{dx}\Big)}\]

Answer 7

how do i simplify to sqrt(1)/2

Answer 8

\[\lim_{x\rightarrow\infty}\frac{\sqrt{\frac {x^2}{x^2}+\frac1{x^2}}}{\frac{2x}x+\frac1x}=\lim_{x\rightarrow\infty}\frac{\sqrt{1+1/x^2}}{2+1/x}\]
\(1/x^2\) and \(1/x\) are almost 0 when x goes to infinity. so only 1in numerator and 2 in denominator are left.

Answer 9

@klimenkov I was actually thinking of squaring the denominator under the radical and then using L'Hopital's. Not from the start

Answer 10

thnaks to all of you

Answer 11

one more if y'all are still here

how to calculate the derivative of f(x)=sinx

Answer 12

i know the derivative of sinx is cosx but i dont know how to prove it with the derivative funtion

Answer 13

\[\lim_{\Delta x\rightarrow0}\frac{\sin(x+\Delta x)-\sin x}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{\sin x \cos\Delta x+\cos x\sin \Delta x-\sin x}{\Delta x}=\]\[\lim_{\Delta x\rightarrow0}\frac{\sin x (\cos\Delta x-1)}{\Delta x}+\lim_{\Delta x\rightarrow0}\frac{\cos x\sin \Delta x}{\Delta x}=\cos x\]
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?

Answer 14

im not exactly sure...

Answer 15

http://www.zweigmedia.com/RealWorld/trig/triglim.html