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anonymous
 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)
anonymous
 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\[\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx2)}{2}=\frac{\cos(0)}{2}=\frac{1}{2}\]

anonymous
 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

anonymous
 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

anonymous
 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 \[\lim_{V \rightarrow 0}\frac{1\cos(V)}{V^2}\] \[=\lim_{V \rightarrow 0}\frac{1\cos(V)}{V^2} \cdot \frac{1+\cos(V)}{1+\cos(V)}\] \[=\lim_{v \rightarrow 0}\frac{1\cos^2(V)}{V^2(1+\cos(V)}=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2(1+\cos(V))}\] \[=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2} \cdot \frac{1}{1+\cos(V))}\] \[=1^2 \cdot \frac{1}{1+\cos(0)}=\frac{1}{1+1}=\frac{1}{2}\]

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