Question

If f(x)={|x|+1, x <0 0, x=0 |x|-1, x>0}
For what value (s) of a does lim x\(\rightarrow\)af(x) exist?

Answer 1

Given:

The function is defined as:

f(x) =

⎧ |x| + 1,    x < 0
⎨ 0,         x = 0
⎩ |x| − 1,    x > 0

---

We are required to find the value(s) of a for which

lim_x→a f(x) exists.

---

Step 1: Consider different cases

Case (i): a < 0

In a neighbourhood of a (where a < 0), f(x) = |x| + 1 = −x + 1.

This is a continuous function.

∴ lim_x→a f(x) exists for all a < 0.

---

Case (ii): a > 0

In a neighbourhood of a (where a > 0), f(x) = |x| − 1 = x − 1.

This is also a continuous function.

∴ lim_x→a f(x) exists for all a > 0.

---

Case (iii): a = 0

Left-hand limit at x = 0:

lim_x→0⁻ f(x) = |0| + 1 = 1

Right-hand limit at x = 0:

lim_x→0⁺ f(x) = |0| − 1 = −1

Since,

lim_x→0⁻ f(x) ≠ lim_x→0⁺ f(x),

the limit at x = 0 does not exist.

---

Conclusion:

The limit lim_x→a f(x) exists for all real values of a except a = 0.

i.e., a ∈ ℝ, a ≠ 0