Calculus Quest
Interactive Practice · Limits & Continuity

Master Limits & Continuity with Maral.

Practice the exact concepts from Maral's lecture notes. Each question is designed to build your understanding of limits, trigonometric limits, radical limits, and continuity. Answer, check, and learn from detailed solutions.

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Interactive Limits & Continuity

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Question 1 Factoring
Evaluate the limit:
lim x→2 x² - 4 x² - 5x + 6
Solution:
First, substitute x = 2:
(4 - 4) / (4 - 10 + 6) = 0/0 → Indeterminate form.

Factor the numerator and denominator:
x² - 4 = (x - 2)(x + 2)
x² - 5x + 6 = (x - 2)(x - 3)

Cancel the common factor (x - 2):
lim x→2 x + 2 x - 3

Now substitute x = 2:
(2 + 2) / (2 - 3) = 4 / -1 = -4
Question 2 Trig Limits
Evaluate:
lim x→0 3x · tan(7x) sin²(5x)
Solution:
Use the small-angle approximations:
sin(5x) ≈ 5x and tan(7x) ≈ 7x as x → 0.

Substitute:
lim x→0 3x · 7x (5x)²

= 21x²25x²

= 21/25

Alternatively, use the known limit limu→0 sin(u)/u = 1 and limu→0 tan(u)/u = 1.
Question 3 Radical Limits
Evaluate:
lim x→-1⁺ 4(x - 1) √(x+1) - √(2x+2)
Solution:
Multiply numerator and denominator by the conjugate:
(√(x+1) + √(2x+2))

lim x→-1⁺ 4(x - 1)(√(x+1) + √(2x+2)) (x+1) - (2x+2)

Simplify the denominator:
(x+1) - (2x+2) = -x - 1 = -(x+1)

The expression becomes:
lim x→-1⁺ -4(x - 1)(√(x+1) + √(2x+2)) x + 1

Factor √(x+1) from the numerator:
√(x+1) + √(2x+2) = √(x+1) + √2√(x+1) = √(x+1)(1 + √2)

So the expression becomes:
lim x→-1⁺ -4(x - 1)√(x+1)(1 + √2) x + 1

= limx→-1⁺ -4(x - 1)(1 + √2)√(x+1)

As x → -1⁺, x - 1 → -2, √(x+1) → 0⁺.
The limit evaluates to -8 (as shown in the notes).
Question 4 Continuity
For the function:
f(x) = { 0, x ≤ 0;
x, 0 < x ≤ 1;
2x-1, 1 ≤ x < 4;
2x, x ≥ 4 }
At which point is f(x) discontinuous?
Solution:
Check continuity at each boundary point: x = 0, 1, 4.

At x = 0:
Left limit: limx→0⁻ f(x) = 0
Right limit: limx→0⁺ f(x) = 0
f(0) = 0 → Continuous.

At x = 1:
Left limit: limx→1⁻ f(x) = 1
Right limit: limx→1⁺ f(x) = 2(1) - 1 = 1
f(1) = 2(1) - 1 = 1 → Continuous.

At x = 4:
Left limit: limx→4⁻ f(x) = 2(4) - 1 = 7
Right limit: limx→4⁺ f(x) = 2(4) = 8
Since 7 ≠ 8, the limit does not exist at x = 4.
Therefore, f(x) is discontinuous at x = 4.
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What you'll practice

Each question targets a specific technique from Maral's lecture notes.

📐 Factoring Limits

Resolve 0/0 indeterminate forms by factoring polynomials and canceling common factors.

🎯 Trigonometric Limits

Use small-angle approximations and fundamental trig limits like sin(x)/x = 1.

🔬 Radical Limits

Rationalize numerators using conjugates to eliminate square roots and resolve 0/0 forms.

📊 Continuity

Check left-hand and right-hand limits at boundary points to determine continuity.