We need to find value of limx→1(1x2+x−2−xx3−1)
=limx→1(1(x−1)(x+2)−x(x−1)(x2+1+x))=limx→1{1(x−1)(1x+2−xx2+x+1)}=limx→1{1(x−1)(x2+x+1−x2−2x(x+2)(x2+x+1))}=limx→1{1(x−1)(1−x(x+2)(x2+x+1))}=limx→1−1(x+2)(x2+x+1)=−1(1+2)(1+1+1)=−19
limx→1(1x2+x−2−xx3−1)
Evaluate the following one sided limits:
(i)limx→2+x−3x2−4
(ii)limx→2−x−3x2−4
(iii)limx→0+13x
(iv)limx→8+2xx+8
(v)limx→0+2x15
(vi)limx→π−2tan x
(vii)limx→π2+sec x
(viii)limx→0−x2−3x+2x3−2x2
(ix)limx→−2+x2−12x+4
(x)limx→0+(2−cot x)
(xi)limx→0−1+cosecx