11st of all, as deno. =0 at x=1, so even numerator shud be=0 at x=1, to make the limit exist finitely.,
so we have p=-4,
if that happens then we also have lim= 2x+3/4x= 5/4 (lopital rule),
so p=-4
cheers!!
f(x)=x2+3x+p /2(x2-1),x≠1 and f(x)=5/4, x=1 , is continues at x=1 then the value of "p" is....??
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4 Answers
1since itz continous at x=1
limx-->1 f(x)=f(x)=5/4
nw u can solve hopefully :P
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