Solutions to exercises on limits and continuity for piecewise functions.

Let’s carefully go through each exercise step by step, checking all calculations and logic before giving the final answer.

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Exercise 1:

Function:
f(x) = { 1 if 0 ≤ x < 1
{ x² if x > 1

Note: The function is not defined at x = 1. There’s no rule for what f(1) equals.

(a) Find limₓ→₁⁺ f(x) and limₓ→₁⁻ f(x)

- Right-hand limit (x → 1⁺): We use the piece for x > 1 → f(x) = x²
So, limₓ→₁⁺ x² = 1² = 1 ✔

- Left-hand limit (x → 1⁻): We use the piece for 0 ≤ x < 1 → f(x) = 1
So, limₓ→₁⁻ 1 = 1 ✔

(b) What does this tell you about limₓ→₁ f(x)?

Since both left and right limits exist and are equal to 1, then:

limₓ→₁ f(x) = 1 ✔

(c) Is f(x) continuous at x = 1?

For continuity at a point, three things must be true:
1. f(a) is defined.
2. limₓ→ₐ f(x) exists.
3. limₓ→ₐ f(x) = f(a)

Here, f(1) is not defined — there’s no value given for x=1 in the piecewise definition.

So even though the limit exists, since f(1) doesn’t exist, the function is not continuous at x=1.

✔ Correct conclusion: No, since f(1) is not defined.

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Exercise 2:

Function:
f(x) = { 1 if 0 ≤ x < 1
{ x² if x ≥ 1

Note: Now, at x = 1, we use the second piece → f(1) = 1² = 1

(a) Find limₓ→₁⁺ f(x) and limₓ→₁⁻ f(x)

- Right-hand limit (x → 1⁺): Use x² → limₓ→₁⁺ x² = 1² = 1 ✔

- Left-hand limit (x → 1⁻): Use 1 → limₓ→₁⁻ 1 = 1 ✔

(b) What does this tell you about limₓ→₁ f(x)?

Left and right limits are both 1 → so limₓ→₁ f(x) = 1 ✔

(c) Is f(x) continuous at x = 1?

Check the three conditions:

1. f(1) is defined? Yes → f(1) = 1² = 1 ✔
2. Limit exists? Yes → it’s 1 ✔
3. Does limit equal f(1)? 1 = 1 → yes ✔

So, YES, f(x) is continuous at x = 1.

The solution says: “Yes, since f(1) = 1² = 1 so f(1) = limₓ→₁ f(x)” → that’s correct.

✔ Final answer for (c): Yes, continuous at x=1.

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Exercise 3:

Function:
f(x) = { 4 - x² if x < 1
{ 2 + x² if x > 1

Note: Again, no definition at x = 1 — only for x < 1 and x > 1.

(a) Find limₓ→₁⁺ f(x) and limₓ→₁⁻ f(x)

- Right-hand limit (x → 1⁺): Use 2 + x² → limₓ→₁⁺ (2 + x²) = 2 + 1² = 3 ✔

- Left-hand limit (x → 1⁻): Use 4 - x² → limₓ→₁⁻ (4 - x²) = 4 - 1² = 3 ✔

(b) What does this tell you about limₓ→₁ f(x)?

Both sides give 3 → so limₓ→₁ f(x) = 3 ✔

(c) Is f(x) continuous at x = 1?

Again, check the three conditions:

1. f(1) defined? → NO! The function has no rule for x=1. Only for x<1 and x>1.

So even though the limit exists (it’s 3), since f(1) is undefined, the function is not continuous at x=1.

✔ Correct conclusion: No, since f(1) is not defined.

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All solutions provided in the image are actually correct.

But let me double-check Exercise 3 part (a):

limₓ→₁⁺ (2 + x²) = 2 + 1 = 3 ✔️
limₓ→₁⁻ (4 - x²) = 4 - 1 = 3 ✔️

And for continuity: missing f(1) → discontinuous ✔️

Same for Exercise 1: same issue — no f(1) → discontinuous ✔️

Exercise 2: f(1) = 1, limit = 1 → continuous ✔️

Everything checks out.

Final Answer: All answers shown in the worksheet are correct as written.