Question

ax^2+bx-2=0 have 2 roots alpha and beta and question is. limx-->beta [(1-cos(ax^2+bx-2)/((ax^2+bx-2)^2)

Answer 1

if x goes to beta, then ax^2+bx-2 goes to 0 since beta is a root of ax^2+bx-2
so thats means we have 0/0 when we plug in beta
we could use l'hospital's rule here
\[\lim_{x \rightarrow \beta}\frac{0-(-\sin(ax^2+bx-2))(2ax+b)}{2(ax^2+bx-2)(2ax+b)}\]

Answer 2

\[\lim_{x \rightarrow \beta}\frac{\sin(ax^2+bx-2)}{2(ax^2+bx-2)}\]

Answer 3

but when we plug in beta we still have 0/0

Answer 4

so we can use l'hosptial again

Answer 5

\[\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx-2)(2ax+b)}{2(2ax+b)}\]

Answer 6

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

Answer 7

wow dude ur answer is right but still have to think how u reach here

Answer 8

do you know l'hospital?

Answer 9

however you spell that dead math guy's name

Answer 10

hmmm ya i heard of it but never used it

Answer 11

so let me show you another way
one sec

Answer 12

ok but i still ddnt get what ever u answer because of that sinx

Answer 13

let \[V(x)=ax^2+bx-2\]
we are given \[V(\beta)=0\]
so if x->beta, then V->0
so we have
\[\lim_{V \rightarrow 0}\frac{1-\cos(V)}{V^2}\]
\[=\lim_{V \rightarrow 0}\frac{1-\cos(V)}{V^2} \cdot \frac{1+\cos(V)}{1+\cos(V)}\]
\[=\lim_{v \rightarrow 0}\frac{1-\cos^2(V)}{V^2(1+\cos(V)}=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2(1+\cos(V))}\]
\[=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2} \cdot \frac{1}{1+\cos(V))}\]
\[=1^2 \cdot \frac{1}{1+\cos(0)}=\frac{1}{1+1}=\frac{1}{2}\]

Answer 14

thanks man this thing really helped me out

Answer 15

thanks girl* :)

Answer 16

lol i am a boy. :P