You’ve learned the concepts — now it’s time to put them to work. This problem set covers every type of limit you’ll see on your Calculus 1 exam, organized from straightforward to challenging. Work through each one, then click Show Solution to check your work.
📚 Table of Contents
- Direct Substitution Problems
- Factoring Problems
- Rationalization Problems
- One-Sided Limit Problems
- Limits at Infinity Problems
- L’Hôpital’s Rule Problems
- Continuity Problems
- Challenge Problems
1. Direct Substitution Problems
For each limit, use direct substitution to find the answer.
Problem 1.
$$\lim_{x \to 3} (4x^2 – 2x + 5)$$
👁️ Show Solution
Plug in x = 3 directly:
$$4(3)^2 – 2(3) + 5 = 36 – 6 + 5 = 35$$
✅ Answer: 35
Problem 2.
$$\lim_{x \to -1} (x^3 + 2x^2 – x + 4)$$
👁️ Show Solution
Plug in x = -1:
$$(-1)^3 + 2(-1)^2 – (-1) + 4 = -1 + 2 + 1 + 4 = 6$$
✅ Answer: 6
Problem 3.
$$\lim_{x \to 2} \frac{3x + 1}{x^2 + 1}$$
👁️ Show Solution
Plug in x = 2:
$$\frac{3(2)+1}{(2)^2+1} = \frac{7}{5}$$
✅ Answer: \(\frac{7}{5}\)
Problem 4.
$$\lim_{x \to 0} (e^x + \cos(x))$$
👁️ Show Solution
Plug in x = 0:
$$e^0 + \cos(0) = 1 + 1 = 2$$
✅ Answer: 2
Problem 5.
$$\lim_{x \to \pi} \sin(x) + 2$$
👁️ Show Solution
Plug in \(x = \pi\):
$$\sin(\pi) + 2 = 0 + 2 = 2$$
✅ Answer: 2
2. Factoring Problems
Each of these gives 0/0 on direct substitution. Factor first, cancel, then substitute.
Problem 6.
$$\lim_{x \to 5} \frac{x^2 – 25}{x – 5}$$
👁️ Show Solution
Direct substitution gives 0/0. Factor the numerator:
$$\lim_{x \to 5} \frac{(x+5)(x-5)}{x-5} = \lim_{x \to 5}(x+5) = 5 + 5 = 10$$
✅ Answer: 10
Problem 7.
$$\lim_{x \to -3} \frac{x^2 + 7x + 12}{x + 3}$$
👁️ Show Solution
Direct substitution gives 0/0. Factor the numerator:
$$\lim_{x \to -3} \frac{(x+3)(x+4)}{x+3} = \lim_{x \to -3}(x+4) = -3 + 4 = 1$$
✅ Answer: 1
Problem 8.
$$\lim_{x \to 2} \frac{x^2 – 3x + 2}{x – 2}$$
👁️ Show Solution
Direct substitution gives 0/0. Factor the numerator:
$$\lim_{x \to 2} \frac{(x-1)(x-2)}{x-2} = \lim_{x \to 2}(x-1) = 2 – 1 = 1$$
✅ Answer: 1
Problem 9.
$$\lim_{x \to 1} \frac{x^3 – 1}{x – 1}$$
👁️ Show Solution
Direct substitution gives 0/0. Use the difference of cubes formula:
$$x^3 – 1 = (x-1)(x^2+x+1)$$
$$\lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{x-1} = \lim_{x \to 1}(x^2+x+1) = 1 + 1 + 1 = 3$$
✅ Answer: 3
Problem 10.
$$\lim_{x \to -2} \frac{x^2 – x – 6}{x + 2}$$
👁️ Show Solution
Direct substitution gives 0/0. Factor the numerator:
$$\lim_{x \to -2} \frac{(x-3)(x+2)}{x+2} = \lim_{x \to -2}(x-3) = -2 – 3 = -5$$
✅ Answer: −5
3. Rationalization Problems
These involve square roots. Multiply by the conjugate to clear the radical, then simplify.
Problem 11.
$$\lim_{x \to 0} \frac{\sqrt{x + 1} – 1}{x}$$
👁️ Show Solution
Direct substitution gives 0/0. Multiply by the conjugate \(\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\):
$$\lim_{x \to 0} \frac{(x+1)-1}{x(\sqrt{x+1}+1)}$$
$$= \lim_{x \to 0} \frac{x}{x(\sqrt{x+1}+1)}$$
$$= \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{1+1} = \frac{1}{2}$$
✅ Answer: \(\frac{1}{2}\)
Problem 12.
$$\lim_{x \to 3} \frac{\sqrt{x + 1} – 2}{x – 3}$$
👁️ Show Solution
Direct substitution gives 0/0. Multiply by the conjugate \(\frac{\sqrt{x+1}+2}{\sqrt{x+1}+2}\):
$$\lim_{x \to 3} \frac{(x+1)-4}{(x-3)(\sqrt{x+1}+2)}$$
$$= \lim_{x \to 3} \frac{x-3}{(x-3)(\sqrt{x+1}+2)}$$
$$= \lim_{x \to 3} \frac{1}{\sqrt{x+1}+2} = \frac{1}{2+2} = \frac{1}{4}$$
✅ Answer: \(\frac{1}{4}\)
Problem 13.
$$\lim_{x \to 0} \frac{\sqrt{4 + x} – 2}{x}$$
👁️ Show Solution
Direct substitution gives 0/0. Multiply by the conjugate \(\frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}\):
$$\lim_{x \to 0} \frac{(4+x)-4}{x(\sqrt{4+x}+2)}$$
$$= \lim_{x \to 0} \frac{x}{x(\sqrt{4+x}+2)}$$
$$= \lim_{x \to 0} \frac{1}{\sqrt{4+x}+2} = \frac{1}{2+2} = \frac{1}{4}$$
✅ Answer: \(\frac{1}{4}\)
Problem 14.
$$\lim_{x \to 5} \frac{\sqrt{x – 1} – 2}{x – 5}$$
👁️ Show Solution
Direct substitution gives 0/0. Multiply by the conjugate \(\frac{\sqrt{x-1}+2}{\sqrt{x-1}+2}\):
$$\lim_{x \to 5} \frac{(x-1)-4}{(x-5)(\sqrt{x-1}+2)}$$
$$= \lim_{x \to 5} \frac{x-5}{(x-5)(\sqrt{x-1}+2)}$$
$$= \lim_{x \to 5} \frac{1}{\sqrt{x-1}+2} = \frac{1}{2+2} = \frac{1}{4}$$
✅ Answer: \(\frac{1}{4}\)
4. One-Sided Limit Problems
For each piecewise function, find the left limit, right limit, and determine whether the two-sided limit exists.
Problem 15. Given
$$f(x) = \begin{cases} 2x + 3 & x < 1 \\ x^2 + 2 & x \geq 1 \end{cases}$$
Find \(\lim_{x \to 1} f(x)\)
👁️ Show Solution
Left limit:
$$\lim_{x \to 1^-}(2x+3) = 2(1)+3 = 5$$
Right limit:
$$\lim_{x \to 1^+}(x^2+2) = 1+2 = 3$$
Since 5 ≠ 3 → ✅ DNE
Problem 16. Given
$$f(x) = \begin{cases} x^2 – 1 & x < 2 \\ 3x - 2 & x \geq 2 \end{cases}$$
Find \(\lim_{x \to 2} f(x)\)
👁️ Show Solution
Left limit:
$$\lim_{x \to 2^-}(x^2-1) = 4-1 = 3$$
Right limit:
$$\lim_{x \to 2^+}(3x-2) = 6-2 = 4$$
Since 3 ≠ 4 → ✅ DNE
Problem 17. Given
$$f(x) = \begin{cases} 4x – 1 & x < 3 \\ x^2 + 2 & x \geq 3 \end{cases}$$
Find \(\lim_{x \to 3} f(x)\)
👁️ Show Solution
Left limit:
$$\lim_{x \to 3^-}(4x-1) = 12-1 = 11$$
Right limit:
$$\lim_{x \to 3^+}(x^2+2) = 9+2 = 11$$
Both equal 11 → ✅ Limit = 11
Problem 18. Given
$$f(x) = \begin{cases} x + 5 & x < 0 \\ 2x + 5 & x \geq 0 \end{cases}$$
Find \(\lim_{x \to 0} f(x)\)
👁️ Show Solution
Left limit:
$$\lim_{x \to 0^-}(x+5) = 5$$
Right limit:
$$\lim_{x \to 0^+}(2x+5) = 5$$
Both equal 5 → ✅ Limit = 5
5. Limits at Infinity Problems
Divide by the highest power of x in the denominator, then evaluate.
Problem 19.
$$\lim_{x \to \infty} \frac{3x^2 + 2x – 1}{x^2 + 5}$$
👁️ Show Solution
Same degree in numerator and denominator → ratio of leading coefficients:
$$\frac{3}{1} = 3$$
✅ Answer: 3
Problem 20.
$$\lim_{x \to \infty} \frac{7x^3 – 4x}{2x^3 + x^2 – 3}$$
👁️ Show Solution
Same degree → ratio of leading coefficients:
$$\frac{7}{2}$$
✅ Answer: \(\frac{7}{2}\)
Problem 21.
$$\lim_{x \to \infty} \frac{5x + 2}{x^2 – 1}$$
👁️ Show Solution
Numerator degree (1) < denominator degree (2) → limit shrinks to zero:
✅ Answer: 0
Problem 22.
$$\lim_{x \to -\infty} \frac{4x^2 – 3}{2x^2 + x}$$
👁️ Show Solution
Same degree → ratio of leading coefficients:
$$\frac{4}{2} = 2$$
✅ Answer: 2
Problem 23.
$$\lim_{x \to \infty} \frac{x^3 + 1}{x^2 – 4}$$
👁️ Show Solution
Numerator degree (3) > denominator degree (2) → grows without bound:
✅ Answer: ∞ (DNE)
6. L’Hôpital’s Rule Problems
Each of these gives 0/0 or ∞/∞. Differentiate the numerator and denominator separately, then evaluate.
Problem 24.
$$\lim_{x \to 0} \frac{\sin(3x)}{x}$$
👁️ Show Solution
Gives 0/0 → Apply L’Hôpital’s Rule:
$$\lim_{x \to 0} \frac{3\cos(3x)}{1} = 3\cos(0) = 3$$
✅ Answer: 3
Problem 25.
$$\lim_{x \to 0} \frac{e^x – 1}{x}$$
👁️ Show Solution
Gives 0/0 → Apply L’Hôpital’s Rule:
$$\lim_{x \to 0} \frac{e^x}{1} = e^0 = 1$$
✅ Answer: 1
Problem 26.
$$\lim_{x \to \infty} \frac{\ln(x)}{x}$$
👁️ Show Solution
Gives ∞/∞ → Apply L’Hôpital’s Rule:
$$\lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0$$
✅ Answer: 0
Problem 27.
$$\lim_{x \to 0} \frac{1 – \cos(x)}{x^2}$$
👁️ Show Solution
Gives 0/0 → Apply L’Hôpital’s Rule:
$$\lim_{x \to 0} \frac{\sin(x)}{2x}$$
Still 0/0 → Apply again:
$$\lim_{x \to 0} \frac{\cos(x)}{2} = \frac{1}{2}$$
✅ Answer: \(\frac{1}{2}\)
Problem 28.
$$\lim_{x \to \infty} \frac{x^3}{e^x}$$
👁️ Show Solution
Gives ∞/∞ → Apply L’Hôpital’s Rule three times:
$$\frac{x^3}{e^x} \to \frac{3x^2}{e^x} \to \frac{6x}{e^x} \to \frac{6}{e^x} = 0$$
✅ Answer: 0
7. Continuity Problems
For each function, determine whether it is continuous at the given point. If not, identify the type of discontinuity.
Problem 29. Is \(f(x) = \frac{x^2 – 9}{x – 3}\) continuous at x = 3?
👁️ Show Solution
f(3) is undefined → condition 1 fails immediately.
The limit exists (equals 6) but f(3) is undefined.
✅ Not continuous — removable discontinuity (hole) at x = 3.
Problem 30. Is \(f(x) = \frac{1}{x – 2}\) continuous at x = 2?
👁️ Show Solution
f(2) is undefined and:
$$\lim_{x \to 2} \frac{1}{x-2} = \pm\infty$$
✅ Not continuous — infinite discontinuity at x = 2.
Problem 31. Given
$$f(x) = \begin{cases} x + 2 & x \neq 1 \\ 5 & x = 1 \end{cases}$$
Is f continuous at x = 1?
👁️ Show Solution
$$\lim_{x \to 1} f(x) = 1 + 2 = 3$$
But f(1) = 5. Since the limit ≠ f(1) → condition 3 fails.
✅ Not continuous — removable discontinuity at x = 1.
Problem 32. Find the value of k that makes
$$f(x) = \begin{cases} kx + 3 & x < 2 \\ x^2 + 1 & x \geq 2 \end{cases}$$
continuous at x = 2.
👁️ Show Solution
For continuity, left limit must equal right limit at x = 2:
$$\lim_{x \to 2^-}(kx+3) = \lim_{x \to 2^+}(x^2+1)$$
$$2k + 3 = 4 + 1 = 5$$
$$2k = 2 \Rightarrow k = 1$$
✅ Answer: k = 1
8. Challenge Problems 🔥
These problems combine multiple techniques. Take your time and think through each step carefully.
Problem 33.
$$\lim_{x \to 0} \frac{\sin(5x)}{\sin(3x)}$$
👁️ Show Solution
Gives 0/0 → Apply L’Hôpital’s Rule:
$$\lim_{x \to 0} \frac{5\cos(5x)}{3\cos(3x)} = \frac{5(1)}{3(1)} = \frac{5}{3}$$
✅ Answer: \(\frac{5}{3}\)
Problem 34.
$$\lim_{x \to \infty} x\sin\!\left(\frac{1}{x}\right)$$
👁️ Show Solution
Let \(t = \frac{1}{x}\), so as \(x \to \infty\), \(t \to 0\):
$$\lim_{t \to 0} \frac{\sin(t)}{t} = 1$$
✅ Answer: 1
Problem 35.
$$\lim_{x \to 0} \frac{e^{2x} – 1}{\sin(x)}$$
👁️ Show Solution
Gives 0/0 → Apply L’Hôpital’s Rule:
$$\lim_{x \to 0} \frac{2e^{2x}}{\cos(x)} = \frac{2(1)}{1} = 2$$
✅ Answer: 2
Problem 36.
$$\lim_{x \to 1} \frac{x^4 – 1}{x^3 – 1}$$
👁️ Show Solution
Gives 0/0 → Factor both numerator and denominator:
$$\lim_{x \to 1} \frac{(x-1)(x^3+x^2+x+1)}{(x-1)(x^2+x+1)} = \frac{1+1+1+1}{1+1+1} = \frac{4}{3}$$
✅ Answer: \(\frac{4}{3}\)
Problem 37.
$$\lim_{x \to \infty} \left(\sqrt{x^2 + x} – x\right)$$
👁️ Show Solution
Multiply by the conjugate \(\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}\):
$$\lim_{x \to \infty} \frac{(x^2+x) – x^2}{\sqrt{x^2+x}+x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2+x}+x}$$
Divide numerator and denominator by x:
$$\lim_{x \to \infty} \frac{1}{\sqrt{1+\frac{1}{x}}+1} = \frac{1}{1+1} = \frac{1}{2}$$
✅ Answer: \(\frac{1}{2}\)
Keep going — every problem you solve is one less surprise on exam day. You’ve got this! 🎓
Found this helpful? Share it with a classmate, and drop a comment below if you want more practice on any specific topic!
