Question

f(x) = |x| + |x − 1| at x = 1

Answer 1

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.