Evaluate the following limits if they exist - Differential Calculus

lim |x − 1| x→1 x2 − 1
LHL=lim |x−1|= lim −(x−1) = lim −1 =−1. x→1− x2 −1 x→1− (x−1)(x+1) x→1− x+1 2
RHL=lim |x−1|= lim (x−1) = lim 1 =1. x→1+ x2 −1 x→1+ (x−1)(x+1) x→1+ x+1 2
Since lim |x−1| ̸= lim |x−1|, the limit lim |x−1| does not exist. x→1− x2−1 x→1+ x2−1 x→1x2−1
(f) (4 marks) lim √xecos(π/2x) x→0+
−1 ≤ cos ( π ) ≤ 1 2x
e−1 ≤ ecos ( π ) ≤ e 2x
√xe−1 ≤ √xecos( π ) ≤ √xe 2x
Now lim √xe−1 =0and lim √xe=0.
x→0+ x→0+ √ cos(π/2x)
Thus by the squeeze theorem, lim xe = 0. x→0+