A community for students.
Here's the question you clicked on:
 0 viewing
nubeer
 4 years ago
ax^2+bx2=0 have 2 roots alpha and beta and question is. limx>beta [(1cos(ax^2+bx2)/((ax^2+bx2)^2)
nubeer
 4 years ago
ax^2+bx2=0 have 2 roots alpha and beta and question is. limx>beta [(1cos(ax^2+bx2)/((ax^2+bx2)^2)

This Question is Closed

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1if x goes to beta, then ax^2+bx2 goes to 0 since beta is a root of ax^2+bx2 so thats means we have 0/0 when we plug in beta we could use l'hospital's rule here \[\lim_{x \rightarrow \beta}\frac{0(\sin(ax^2+bx2))(2ax+b)}{2(ax^2+bx2)(2ax+b)}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow \beta}\frac{\sin(ax^2+bx2)}{2(ax^2+bx2)}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1but when we plug in beta we still have 0/0

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so we can use l'hosptial again

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx2)(2ax+b)}{2(2ax+b)}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx2)}{2}=\frac{\cos(0)}{2}=\frac{1}{2}\]

nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0wow dude ur answer is right but still have to think how u reach here

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1do you know l'hospital?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1however you spell that dead math guy's name

nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0hmmm ya i heard of it but never used it

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so let me show you another way one sec

nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0ok but i still ddnt get what ever u answer because of that sinx

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1let \[V(x)=ax^2+bx2\] we are given \[V(\beta)=0\] so if x>beta, then V>0 so we have \[\lim_{V \rightarrow 0}\frac{1\cos(V)}{V^2}\] \[=\lim_{V \rightarrow 0}\frac{1\cos(V)}{V^2} \cdot \frac{1+\cos(V)}{1+\cos(V)}\] \[=\lim_{v \rightarrow 0}\frac{1\cos^2(V)}{V^2(1+\cos(V)}=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2(1+\cos(V))}\] \[=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2} \cdot \frac{1}{1+\cos(V))}\] \[=1^2 \cdot \frac{1}{1+\cos(0)}=\frac{1}{1+1}=\frac{1}{2}\]

nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0thanks man this thing really helped me out
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.