A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
would lim(x approches infinity) (sqrt(x^2+1))/(2x+1) be DNE since it would basically be infinity/infinity?
anonymous
 4 years ago
would lim(x approches infinity) (sqrt(x^2+1))/(2x+1) be DNE since it would basically be infinity/infinity?

This Question is Closed

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.3@redham L'hopital's is not the best idea.

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

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

klimenkov
 4 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.

anonymous
 4 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

anonymous
 4 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

anonymous
 4 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
 4 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im not exactly sure...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.