## anonymous 4 years ago would lim(x approches infinity) (sqrt(x^2+1))/(2x+1) be DNE since it would basically be infinity/infinity?

1. klimenkov

What is DNE?

2. anonymous

does not exist

3. 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

4. klimenkov

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

5. klimenkov

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

6. Valpey

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

7. anonymous

how do i simplify to sqrt(1)/2

8. 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.

9. anonymous

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

10. anonymous

thnaks to all of you

11. anonymous

one more if y'all are still here how to calculate the derivative of f(x)=sinx

12. anonymous

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

13. 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?

14. anonymous

im not exactly sure...

15. Valpey