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

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

thanks man this thing really helped me out

thanks girl* :)

lol i am a boy. :P