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Question
f(x) = |x| + |x − 1| at x = 1
Solution
We have, f(x) = |x| + |x − 1| at x = 1
At x = 1
L.H.L. = `lim_(x -> 1^-) [|x| + |x - 1|]`
= `lim_("h" -? 0^-) [|1 - "h"| + |1 - "h" - 1|]`
= 1 + 0
= 1
And R.H.L. = `lim_(x ->^+) [|x| + x - 1|]`
= `lim_("h" -> 0) [|1 + "h"| + |1 + "h" - 1|]`
= 1 + 0
= 1
Also f(1) = |1| + |0| = 1
Thus, L.H.L. = R.H.L = f(1)
Hence, f(x) is continuous at x = 1
Alternative method:
Since every modulus function is continuous for all real x
|x| and |x – 1| are continuous for all real x.
So, |x| + |x – 1| is continuous for all real x and hence at x = 0.
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