## nubeer Group Title 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) 3 years ago 3 years ago

1. myininaya

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 $\lim_{x \rightarrow \beta}\frac{0-(-\sin(ax^2+bx-2))(2ax+b)}{2(ax^2+bx-2)(2ax+b)}$

2. myininaya

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

3. myininaya

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

4. myininaya

so we can use l'hosptial again

5. myininaya

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

6. myininaya

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

7. nubeer

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

8. myininaya

do you know l'hospital?

9. myininaya

however you spell that dead math guy's name

10. nubeer

hmmm ya i heard of it but never used it

11. myininaya

so let me show you another way one sec

12. nubeer

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

13. myininaya

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

14. nubeer

thanks man this thing really helped me out

15. myininaya

thanks girl* :)

16. nubeer

lol i am a boy. :P