NCERT Solutions
Class 11 Maths
Limits and Derivatives

Ex.13.1 Q.30
If f(x) =|x| + 1, {if x < 0
If f(x) = 0, {if x = 0
If f(x) = |x| - 1, {if x > 0
For what value(s) of a does limx->a f(x) exists?
Given function is
|x| + 1, if x < 0 f(x) = 0, if x = 0
|x| - 1, if x > 0
When a = 0,
lim x ->0- f(x) = lim x -> 0- [|x| + 1]
= lim x -> 0 (-x + 1)
[When x < 0, |x| = -x]
= 1 lim x ->0+ f(x) = lim x -> 0+ [|x| - 1]
= lim x -> 0 (x - 1)
[When x > 0, |x| = x]
= -1
Since lim x ->0- f(x) ≠ lim x ->0+ f(x) = 0
Hence, lim x -> 0 f(x) = 0 When a < 0
lim x ->a- f(x) = lim x -> a- [|x| + 1]
= lim x -> a (-x + 1)
[When x < a < 0, |x| = -x]
= -a + 1 lim x >a+ f(x) = lim x -> a+ [|x| + 1]
= lim x -> a (-x + 1)
[When a < x < 0, |x| = -x]
= -a + 1
Since lim x ->a- f(x) = lim x ->a+ f(x) = -a + 1
Thus, lim x -> a f(x) exists at x= 0 where a < 0 When a > 0
lim x ->a- f(x) = lim x -> a- [|x| - 1]
= lim x -> a (x - 1)
[When 0 < x < a, |x| = x] = a + 1 lim x ->a+ f(x) = lim x -> a+ [|x| - 1]
= lim x -> a (x - 1)
[When 0 < a < x, |x| = x]
= a - 1
Since lim x ->a- f(x) = lim x ->a+ f(x) = a - 1 Thus, lim x -> a f(x) exists at x= 0 where a > 0