If f (x) = {|x| + 1, x < 0; 0, x = 0; and |x| - 1, x > 0}
For what value (s) of a does limₓ→ₐ f (x) exists?

Solution:

The given function is f (x) = {|x| + 1, x < 0, 0, x = 0 and |x| - 1, x > 1}

We will evaluate the left hand and the right hand limits when a = 0.

limₓ→₀₋ f (x) = limₓ→₀₋ [|x| + 1]

= limₓ→₀ (- x + 1) [When x < 0, |x| = - x]

= 0 + 1

= 1

limₓ→₀₊ f (x) = limₓ→₀₊ [|x| - 1]

= limₓ→₀ [x - 1] [∵ If x > 0, |x| = x]

= 0 - 1

= - 1

Here it is observed that limₓ→₀₋ f (x) ≠ limₓ→₀₊ f (x)

Therefore, limₓ→₀ f (x) does not exist at x = a where a = 0.

Now we will see the existence of this limit when a < 0 and when a > 0.

When a < 0:

limₓ→ₐ₋ f (x) = limₓ→ₐ₋ [|x| + 1]

= limₓ→ₐ (- x + 1) [x < a < 0, ⇒ |x| = - x]

= - a + 1

limₓ→ₐ₊ f (x) = limₓ→ₐ₊ [|x| + 1]

= limₓ→ₐ [- x + 1] [∵ a < x < 0 ⇒ |x| = -x]

= - a + 1

Hence, limₓ→ₐ₋ f (x) = limₓ→ₐ₊ f (x) = - a + 1

Thus, limit of f (x) exists at x = a where a < 0.

When a > 0:

limₓ→ₐ₋ f (x) = limₓ→ₐ₋ [|x| + 1]

= limₓ→ₐ (x - 1) [∵ When x > a > 0, |x| = x]

= a - 1

limₓ→ₐ₊ f (x) = limₓ→ₐ₊ [|x| - 1]

= limₓ→ₐ [x - 1] [0 < a < x ⇒ |x| = x]

= a - 1

Hence, limₓ→ₐ₋ f (x) = limₓ→ₐ₊ f (x) = a - 1

Thus, limit of f (x) exists at x = a where a > 0.

Hence, limₓ→ₐ f (x) exists for all a ≠ 0

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NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.1 Question 30

If f (x) = {|x| + 1, x < 0, 0, x = 0 and |x| - 1, x > 1} For what value (s) of does limₓ→ₐ f (x) exists?

Summary:

The limit limₓ→ₐ f (x) exists for all a ≠ 0