solve the limits question given in attachment ..................................................................

lim x tends to 1 (ax^2+bx+c)/(x – 1)^2 = 2

the denominator is tending to zero. so for this lim to exist, the numerator should also tend to zero else the limit would become infinite. so a(1)^2+b*1+c=0 or a+b+c=0 or c= -a-b

so we substitute value of c in (ax^2+bx+c)/(x – 1)^2 to get (ax^2+bx+c)/(x – 1)^2 = [a(x+1)+b]/(x-1). now for this lim to exist, numerator again should be zero since denom tends to zero.

so a(1+1)+b=0 or 2a+b=0.

so lim becomes  [a(x+1)+b]/(x-1)=  [a(x+1)-2a]/(x-1)= a

so a=2. similarly find b and c, and to find the limit (x-a)(x-b)(x-c)/(x+1) plug in those values and put x=1