Limits & Continuity Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice limits and continuity with the most important tools you need for Calculus: limit notation \(\lim_{x\to a} f(x)\) and the meaning of "approach," direct substitution for continuous functions (polynomials, trig, exponentials), core limit laws (sum, product, quotient, constant multiple), indeterminate forms like \(0/0\) and how to fix them with factoring and canceling, rationalizing with conjugates for radicals, the must-know special limits \(\lim_{x\to 0}\dfrac{\sin x}{x}=1\) and \(\lim_{x\to 0}\dfrac{e^x-1}{x}=1\), limits at infinity for rational functions (degrees, leading coefficients, horizontal asymptotes), one-sided limits \(\lim_{x\to a^-}\) and \(\lim_{x\to a^+}\), and continuity tests, including checking piecewise functions at break points. If you want a refresher, click Start lesson to open a step-by-step guide with worked examples and quick checks.
Answer the question set and review your mistakes at the end.
How this limits and continuity practice works
1. Take the practice set: answer the limits and continuity questions below.
2. Open the lesson (optional): review limit laws, special limits, limits at infinity, one-sided limits, and continuity with clear examples.
3. Retry: return to the question set and apply the limit rules and continuity conditions immediately.
What you will learn in the limits & continuity lesson
Limit basics & direct substitution
Limit notation \(\lim_{x\to a} f(x)\) and the "approach" idea
Direct substitution for continuous functions: polynomials, trig, exponentials
Purpose: Build a clear understanding of limits and continuity so you can evaluate limits of functions using limit laws and direct substitution, handle indeterminate forms like \(0/0\) using factoring and canceling or rationalizing, use the key special limits \(\lim_{x\to 0}\dfrac{\sin x}{x}=1\) and \(\lim_{x\to 0}\dfrac{e^x-1}{x}=1\), compute limits at infinity for rational functions, and test continuity (including piecewise functions, one-sided limits, and common discontinuities).
Success criteria
Interpret limit notation \(\lim_{x\to a} f(x)\) and the meaning of “approach.”
Evaluate limits of constants, polynomials, trig functions, and exponentials by direct substitution when the function is continuous.
Use the basic limit laws (sum, product, quotient, constant multiple).
Fix indeterminate forms like \(0/0\) using factoring and canceling common factors.
Use rationalizing (conjugates) to simplify limits with radicals.
Use the special limits \(\lim_{x\to 0}\dfrac{\sin x}{x}=1\) and \(\lim_{x\to 0}\dfrac{e^x-1}{x}=1\) (with angles in radians).
Compute limits at infinity for rational functions and identify horizontal asymptotes.
Compute one-sided limits and decide when a two-sided limit exists.
Check continuity at a point using \(\lim_{x\to a} f(x)=f(a)\).
Recognize removable, jump, and infinite discontinuities (vertical asymptotes).
Key vocabulary
Limit: the value \(f(x)\) approaches as \(x\) approaches a point \(a\).
Left-hand limit: \(\lim_{x\to a^-} f(x)\).
Right-hand limit: \(\lim_{x\to a^+} f(x)\).
Indeterminate form: an algebraic form like \(0/0\) that requires simplification before taking a limit.
Limit at infinity: \(\lim_{x\to\infty} f(x)\) or \(\lim_{x\to-\infty} f(x)\); often used to find asymptotes.
Continuity at \(a\): \(f\) is continuous at \(a\) if \(f(a)\) exists, \(\lim_{x\to a} f(x)\) exists, and they are equal.
Pre-check 1: What is \(\displaystyle \lim_{x \to -2} 7\)?
Hint: The limit of a constant is the constant.
Pre-check 2: Does \(\displaystyle \lim_{x \to 0} \frac{|x|}{x}\) exist?
Hint: For \(x>0\), \(|x|/x=1\). For \(x<0\), \(|x|/x=-1\).
Limit Basics
Limits, direct substitution, and the basic limit laws
Learning goal: Evaluate common limits quickly using direct substitution and the main limit laws.
Key idea
A limit describes what value a function approaches as the input approaches a number: \[ \lim_{x\to a} f(x). \] If \(f\) is continuous at \(a\), then the limit is found by direct substitution: \[ \lim_{x\to a} f(x)=f(a). \] This works for polynomials, and also for rational functions as long as the denominator is not zero at \(a\).
Limit laws you’ll use constantly
Sum: \(\lim (f+g)=\lim f+\lim g\)
Product: \(\lim (fg)=(\lim f)(\lim g)\)
Quotient: \(\lim \frac{f}{g}=\frac{\lim f}{\lim g}\) if \(\lim g≠ 0\)
Because \(x^2+1\) is a polynomial, it is continuous everywhere, so substitute \(x=3\): \[ \lim_{x\to 3}(x^2+1)=3^2+1=9+1=10. \]
Try it
Try it 1: What is \(\displaystyle \lim_{x \to 5} (x - 2)\)?
Hint: Substitute \(x=5\).
Try it 2: What is \(\displaystyle \lim_{x \to \pi} \sin(x)\)?
Hint: \(\sin(x)\) is continuous, so substitute \(x=\pi\).
Summary
For continuous functions, \(\lim_{x\to a} f(x)=f(a)\).
Limit laws let you break complicated limits into simpler ones.
Algebraic Limits
Indeterminate forms \(0/0\): factoring, canceling, and rationalizing
Learning goal: When substitution gives \(0/0\), simplify first — then evaluate the limit.
Key idea
If direct substitution produces an indeterminate form like \(\frac{0}{0}\), the limit is not found yet. Instead, simplify the expression (without plugging in) and then take the limit. Two common tools:
Factor and cancel: factor the numerator/denominator and cancel a common factor (after factoring).
Rationalize: multiply by a conjugate to remove radicals like \(\sqrt{x^2+1}-x\).
\(\lim_{x\to 0}\dfrac{\sin x}{x}=1\) and \(\lim_{x\to 0}\dfrac{e^x-1}{x}=1\).
Use substitutions and scaling to match the special-limit forms.
Limits at Infinity
Limits at infinity for rational functions and asymptotes
Learning goal: Use dominant terms to compute limits at infinity and identify horizontal asymptotes.
Key idea
For rational functions \(\dfrac{P(x)}{Q(x)}\), compare the degrees (highest powers) of \(P\) and \(Q\). A quick rule for \(\lim_{x\to\infty}\dfrac{P(x)}{Q(x)}\):
If \(\deg(P) < \deg(Q)\), the limit is \(0\).
If \(\deg(P) = \deg(Q)\), the limit is the ratio of leading coefficients.
If \(\deg(P) > \deg(Q)\), the expression grows without bound (often \(\pm\infty\)), so there is no horizontal asymptote.
The degrees are equal (both are \(3\)), so divide by \(x^3\): \[ \frac{2x^3+1}{x^3-2}=\frac{2+\frac{1}{x^3}}{1-\frac{2}{x^3}}. \] As \(x\to\infty\), \(\frac{1}{x^3}\to 0\) and \(\frac{2}{x^3}\to 0\), so \[ \lim_{x\to\infty}\frac{2x^3+1}{x^3-2}=\frac{2+0}{1-0}=2. \]
Try it
Try it 1: What is \(\displaystyle \lim_{x \to \infty} \frac{5}{x}\)?
Hint: A constant divided by a growing \(x\) goes to \(0\).
Try it 2: What is \(\displaystyle \lim_{x\to\infty} \frac{3x^3 - x}{x^3 + 2}\)?
Hint: Divide numerator and denominator by \(x^3\) and keep only the leading coefficients.
Summary
At infinity, compare degrees: smaller degree on top \(\Rightarrow 0\); equal degrees \(\Rightarrow\) ratio of leading coefficients.
These limits often tell you the horizontal asymptote \(y=L\).
Continuity
Continuity at a point and continuity of piecewise functions
Learning goal: Use the definition of continuity and one-sided limits to check piecewise functions.
Key idea
A function \(f\) is continuous at \(x=a\) if all three conditions hold:
\(f(a)\) is defined,
\(\lim_{x\to a} f(x)\) exists,
\(\lim_{x\to a} f(x)=f(a)\).
The two-sided limit \(\lim_{x\to a} f(x)\) exists exactly when the one-sided limits match: \[ \lim_{x\to a^-} f(x)=\lim_{x\to a^+} f(x). \]
Worked example
Example: Is \(f(x)=\begin{cases}x^2, & x < 1\\2x-1, & x\ge 1\end{cases}\) continuous at \(x=1\)?
Left-hand limit: \[ \lim_{x\to 1^-} x^2 = 1. \] Right-hand limit: \[ \lim_{x\to 1^+} (2x-1)=2(1)-1=1. \] So \(\lim_{x\to 1} f(x)\) exists and equals \(1\). Also \(f(1)=2(1)-1=1\). Therefore \(f\) is continuous at \(x=1\).
Try it
Try it 1: Is \(f(x)=\begin{cases}x^2, & x < 1\\2x-1, & x\ge 1\end{cases}\) continuous at \(x=1\)?
Hint: Compute \(\lim_{x\to 1^-}x^2\), \(\lim_{x\to 1^+}(2x-1)\), and compare to \(f(1)\).
Try it 2: Is \(f(x)=|x+1|\) continuous at \(x=-1\)?
Hint: \(|x+1|\) is continuous for all real \(x\), including \(x=-1\).
Summary
Continuity at \(a\): \(f(a)\) exists, \(\lim_{x\to a}f(x)\) exists, and they are equal.
For piecewise functions, check left and right limits at the break point.
Discontinuities
When limits fail: jump discontinuities, infinite limits, and DNE
Learning goal: Use one-sided limits to decide whether a limit exists, and recognize common discontinuities.
Key idea
A two-sided limit \(\lim_{x\to a} f(x)\) exists only if the left and right limits are equal. If they are different, the limit does not exist (often a jump discontinuity). If the function grows without bound (toward \(\pm\infty\)), you have an infinite discontinuity (a vertical asymptote).
Worked example
Example: Does \(\displaystyle \lim_{x\to 0} \frac{|x|}{x}\) exist?
For \(x>0\), \(\frac{|x|}{x}=1\). For \(x<0\), \(\frac{|x|}{x}=-1\). So \[ \lim_{x\to 0^-}\frac{|x|}{x}=-1,\quad \lim_{x\to 0^+}\frac{|x|}{x}=1. \] Because the one-sided limits are different, the two-sided limit does not exist.
Try it
Try it 1: Does \(\displaystyle \lim_{x \to 0} \frac{1}{x}\) exist?
Hint: As \(x\to 0^+\), \(1/x\to +\infty\). As \(x\to 0^-\), \(1/x\to -\infty\).
Try it 2: What is \(\displaystyle \lim_{x \to 1^+} \frac{1}{x-1}\)?
Hint: Approaching \(1\) from the right makes \(x-1\) a very small positive number.
Summary
If left \(≠\) right, the two-sided limit does not exist (jump-type behavior).
If the function blows up to \(\pm\infty\), you have an infinite limit and a vertical asymptote.
Applications & Big Picture
Why limits and continuity matter
Learning goal: Connect limits and continuity to the next big topics in calculus — and finish with a final check.
Where limits and continuity show up
Derivatives: the derivative is defined using a limit of a difference quotient.
Integrals: area and accumulation are built from limits of sums.
Graphs: continuity explains when a graph can be drawn without lifting your pencil.
Modeling: physics, economics, and biology use limits to describe “instantaneous” behavior and long-run trends.
Worked example: filling a hole to make a function continuous
Example: Let \(g(x)=\dfrac{x^3-1}{x-1}\) for \(x≠ 1\). Find \(\displaystyle \lim_{x\to 1} g(x)\) and choose \(g(1)\) so that \(g\) becomes continuous at \(x=1\).
Factor \(x^3-1=(x-1)(x^2+x+1)\), so for \(x≠ 1\), \[ g(x)=x^2+x+1. \] Then \[ \lim_{x\to 1} g(x)=1^2+1+1=3. \] To make \(g\) continuous at \(x=1\), define \(g(1)=3\).
Try it
Try it 1: What is \(\displaystyle \lim_{x \to 1} \frac{x^3 - 1}{x - 1}\)?
Hint: Factor \(x^3-1=(x-1)(x^2+x+1)\), cancel, then substitute.
Try it 2: What is \(\displaystyle \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n\)?
Hint: This is the classic definition of the constant \(e\).
Final recap
Direct substitution: if \(f\) is continuous at \(a\), then \(\lim_{x\to a} f(x)=f(a)\).
0/0 limits: simplify first (factor/cancel or rationalize), then substitute.
Special limits: \(\lim_{x\to 0}\dfrac{\sin x}{x}=1\) and \(\lim_{x\to 0}\dfrac{e^x-1}{x}=1\).
Limits at infinity: compare degrees or divide by the highest power of \(x\).
One-sided limits: a two-sided limit exists only if left and right match.
Continuity: \(f(a)\) defined, limit exists, and \(\lim_{x\to a} f(x)=f(a)\).
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the limit or continuity skill you need.
Practice set
Limits & Continuity practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
What is \(\lim_{x \to 3} 5\)?
Correct answer: C. \(5\)
Explanation: A constant limit equals the constant: \(5\).
Question 2Not answered
Is the function f(x)=\begin{cases}x^2 & x≠1\\1 & x=1\end{cases} continuous at \(x=1\)?
Correct answer: A. Continuous
Explanation: The limit as \(x\to1\) of \(x^2\) is \(1\) and \(f(1)=1\), so it is continuous at \(1\).
Question 3Not answered
What is \(\lim_{x \to 2} (3x + 1)\)?
Correct answer: B. \(7\)
Explanation: Substitute \(x=2\): \(3·2+1=7\).
Question 4Not answered
What is \(\lim_{x \to 0} \dfrac{\sin(x)}{x}\)?
Correct answer: C. \(1\)
Explanation: Standard limit: \(1\).
Question 5Not answered
Does \(\lim_{x \to 0} \tfrac{1}{x}\) exist?
Correct answer: C. Does not exist
Explanation: As \(x\to0\), \(1/x\) → \(+\infty\) from right and \(-\infty\) from left, so no.
Question 6Not answered
What is \(\lim_{x \to 0^+} \tfrac{1}{x}\)?
Correct answer: A. \(+\infty\)
Explanation: As \(x\to0^+\), \(1/x\to+\infty\).
Question 7Not answered
What is \(\lim_{x \to 4} \dfrac{x-4}{x-4}\)?
Correct answer: D. 1
Explanation: Simplify for x≠4: quotient = 1.
Question 8Not answered
Is \(f(x)=|x|\) continuous at \(x=0\)?
Correct answer: C. Continuous
Explanation: Both left and right limits = 0 and \(f(0)=0\), so yes.
Question 9Not answered
Does f(x)=\begin{cases}\frac{x^2-1}{x-1}&x≠1\\2&x=1\end{cases} have a removable discontinuity at 1?
Correct answer: C. No
Explanation: \((x^2-1)/(x-1)=x+1\), so \(\lim_{x\to1}f(x)=2=f(1)\); the function is continuous at 1, so there is no removable discontinuity.
