Here's the question you clicked on:
nubeer
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)
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 \[\lim_{x \rightarrow \beta}\frac{0-(-\sin(ax^2+bx-2))(2ax+b)}{2(ax^2+bx-2)(2ax+b)}\]
\[\lim_{x \rightarrow \beta}\frac{\sin(ax^2+bx-2)}{2(ax^2+bx-2)}\]
but when we plug in beta we still have 0/0
so we can use l'hosptial again
\[\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx-2)(2ax+b)}{2(2ax+b)}\]
\[\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx-2)}{2}=\frac{\cos(0)}{2}=\frac{1}{2}\]
wow dude ur answer is right but still have to think how u reach here
do you know l'hospital?
however you spell that dead math guy's name
hmmm ya i heard of it but never used it
so let me show you another way one sec
ok but i still ddnt get what ever u answer because of that sinx
let \[V(x)=ax^2+bx-2\] we are given \[V(\beta)=0\] so if x->beta, then V->0 so we have \[\lim_{V \rightarrow 0}\frac{1-\cos(V)}{V^2}\] \[=\lim_{V \rightarrow 0}\frac{1-\cos(V)}{V^2} \cdot \frac{1+\cos(V)}{1+\cos(V)}\] \[=\lim_{v \rightarrow 0}\frac{1-\cos^2(V)}{V^2(1+\cos(V)}=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2(1+\cos(V))}\] \[=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2} \cdot \frac{1}{1+\cos(V))}\] \[=1^2 \cdot \frac{1}{1+\cos(0)}=\frac{1}{1+1}=\frac{1}{2}\]
thanks man this thing really helped me out