Limits

Understand limits as nearby behavior and learn step-by-step limit workflows.

A limit describes what a function’s outputs approach as the input gets close to a target value. Limits are the foundation of derivatives, integrals, and much of calculus because they let us reason about values that are approached, even when a formula is undefined at one point.

The most important beginner habit is this:

A limit is about nearby behavior, not only the value at the target point.

Learning objectives

After this lesson you should be able to:

  • explain a limit as a value a function approaches;
  • estimate limits from tables and graphs;
  • distinguish a function value from a limiting value;
  • use direct substitution when appropriate;
  • evaluate removable-hole limits by factoring and canceling;
  • rationalize simple radical limits;
  • compare left-hand and right-hand limits;
  • identify simple infinite limits and vertical asymptotes;
  • check limits with SymPy.

Vocabulary and notation

The notation

\[ \lim_{x\to a} f(x)=L \]

means: as \(x\) gets close to \(a\), the output \(f(x)\) gets close to \(L\).

Important words:

  • approaches: gets close to, not necessarily equals;
  • target value: the input value \(a\);
  • limiting value: the output value \(L\);
  • left-hand limit: approach from values less than \(a\);
  • right-hand limit: approach from values greater than \(a\).

Limit problem-solving roadmap

Use this order for most beginner limit problems.

  1. Try direct substitution. Put the target value into the expression.
  2. Interpret the result.
    • A normal number usually gives the limit.
    • 0/0 means more work is needed.
    • nonzero divided by zero suggests one-sided or infinite behavior.
  3. Choose a repair method. Factor, cancel, rationalize, or check one-sided behavior.
  4. Simplify carefully. Only cancel common factors, not terms.
  5. Substitute again. Use the simplified expression when it matches the original nearby.
  6. State the result in words. Say what the outputs approach.

Direct substitution

If the function is continuous at the target value, direct substitution works. Polynomials are continuous everywhere, so limits of polynomial expressions can be found by substituting.

Walkthrough: direct substitution

Evaluate

\[ \lim_{x\to 2}(x^2+3x). \]

  1. Read the problem. We need the output approached as \(x\) gets close to 2.
  2. Classify the function. This is a polynomial, so direct substitution is allowed.
  3. Substitute. \[ 2^2+3(2)=4+6=10. \]
  4. Check. Values near 2, such as 1.9 and 2.1, produce outputs near 10.
  5. Interpret. As \(x\) approaches 2, \(x^2+3x\) approaches 10.

So

\[ \lim_{x\to 2}(x^2+3x)=10. \]

Guided exercise: direct substitution

Evaluate \(\lim_{x\to -1}(x^2-4x+1)\).

  1. What type of expression is it?
  2. What value should you substitute?
  3. Compute the output.
  4. State the limit in a sentence.
  1. It is a polynomial, so direct substitution works.
  2. Substitute \(x=-1\).
  3. \((-1)^2-4(-1)+1=1+4+1=6\).
  4. As \(x\) approaches \(-1\), the function approaches 6.

Limits from tables

A table helps when a graph or formula is hard to read. Choose input values close to the target from both sides.

For

\[ f(x)=\frac{x^2-1}{x-1}, \]

direct substitution at \(x=1\) gives \(0/0\). But nearby values suggest a limit:

\(x\) \(f(x)\)
0.9 1.9
0.99 1.99
1.01 2.01
1.1 2.1

The outputs approach 2. The table is evidence; algebra explains why.

Removable holes: factor and cancel

When direct substitution gives 0/0, the expression may have a removable hole. That does not mean the limit is zero. It means the original formula needs to be simplified for nearby values.

Walkthrough: factor and cancel

Evaluate

\[ \lim_{x\to 3}\frac{x^2-9}{x-3}. \]

  1. Try substitution. \[ \frac{3^2-9}{3-3}=\frac{0}{0}. \] This is indeterminate, so we do not stop.
  2. Factor. \[ x^2-9=(x-3)(x+3). \]
  3. Rewrite the expression. \[ \frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}. \]
  4. Cancel the common factor for \(x\ne 3\). \[ \frac{(x-3)(x+3)}{x-3}=x+3. \] We are allowed to use this simplified expression for nearby values, even though the original expression is undefined exactly at \(x=3\).
  5. Substitute again. \[ 3+3=6. \]
  6. Interpret. The original function has a hole at \(x=3\), but nearby outputs approach 6.

Therefore,

\[ \lim_{x\to 3}\frac{x^2-9}{x-3}=6. \]

WarningCommon mistake

Do not cancel terms that are added or subtracted. You may cancel common factors. For example, in \(\frac{(x-3)(x+3)}{x-3}\), the factor \(x-3\) is common. In \(\frac{x^2-9}{x-3}\), you must factor first.

Guided exercise: another factor-and-cancel limit

Evaluate

\[ \lim_{x\to 2}\frac{x^2-5x+6}{x-2}. \]

  1. What happens with direct substitution?
  2. Factor the numerator.
  3. Cancel the common factor.
  4. Substitute into the simplified expression.
  1. Direct substitution gives \((4-10+6)/(0)=0/0\), so more work is needed.
  2. \(x^2-5x+6=(x-2)(x-3)\).
  3. For \(x\ne 2\), \[ \frac{(x-2)(x-3)}{x-2}=x-3. \]
  4. Substitute \(x=2\): \(2-3=-1\).

The limit is \(-1\).

Radical limits: rationalize

Some 0/0 limits contain square roots. Multiplying by the conjugate can reveal a canceling factor.

Walkthrough: rationalizing

Evaluate

\[ \lim_{x\to 0}\frac{\sqrt{x+1}-1}{x}. \]

  1. Try substitution. \[ \frac{\sqrt{1}-1}{0}=\frac{0}{0}. \]
  2. Multiply by the conjugate. \[ \frac{\sqrt{x+1}-1}{x}\cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}. \]
  3. Use difference of squares in the numerator. \[ (\sqrt{x+1}-1)(\sqrt{x+1}+1)=(x+1)-1=x. \]
  4. Simplify. \[ \frac{x}{x(\sqrt{x+1}+1)}=\frac{1}{\sqrt{x+1}+1}. \]
  5. Substitute. \[ \frac{1}{\sqrt{1}+1}=\frac12. \]

So the limit is \(\frac12\).

One-sided limits

A two-sided limit exists only when the left-hand and right-hand limits agree.

Walkthrough: compare sides

Suppose a piecewise function has this behavior near \(x=1\):

\[ \lim_{x\to 1^-}f(x)=2, \qquad \lim_{x\to 1^+}f(x)=5. \]

  1. The left-hand approach gives 2.
  2. The right-hand approach gives 5.
  3. These values do not match.
  4. Therefore the two-sided limit does not exist.

If both one-sided limits were 2, then the two-sided limit would be 2.

Infinite limits and vertical asymptotes

For \(f(x)=1/x\) near \(x=0\):

  • as \(x\to 0^+\), outputs grow positive without bound;
  • as \(x\to 0^-\), outputs grow negative without bound.

So the two-sided finite limit does not exist. The graph has a vertical asymptote at \(x=0\).

For \(f(x)=1/x^2\) near \(x=0\), both sides grow positive without bound. We often write

\[ \lim_{x\to 0}\frac1{x^2}=\infty, \]

but this means the outputs grow without bound, not that infinity is an ordinary number.

Common mistakes and repairs

Mistake 1: stopping at 0/0

Repair: say “direct substitution is inconclusive” and try factoring, rationalizing, or another method.

Mistake 2: confusing \(f(a)\) with the limit

Repair: remember that a hole can make \(f(a)\) undefined while the limit still exists.

Mistake 3: ignoring one-sided behavior

Repair: when a formula changes, a denominator becomes zero, or a graph jumps, check from the left and right separately.

Mixed practice

  1. Evaluate \(\lim_{x\to 4}(x^2-2x)\).
  2. Evaluate \(\lim_{x\to 4}\frac{x^2-16}{x-4}\).
  3. Evaluate \(\lim_{x\to 0}\frac{\sqrt{x+9}-3}{x}\).
  4. If the left-hand limit is -1 and the right-hand limit is -1, what is the two-sided limit?
  5. If the left-hand limit is -1 and the right-hand limit is 1, does the two-sided limit exist?
  6. Explain why \(\lim_{x\to 0}1/x\) does not exist as a two-sided finite limit.
  1. Direct substitution gives \(16-8=8\).
  2. Factor: \((x^2-16)/(x-4)=((x-4)(x+4))/(x-4)=x+4\), so the limit is 8.
  3. Multiply by the conjugate: \[ \frac{\sqrt{x+9}-3}{x}\cdot\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} =\frac{1}{\sqrt{x+9}+3}. \] Substitute \(x=0\) to get \(1/6\).
  4. The two-sided limit is -1.
  5. No. The one-sided limits disagree.
  6. From the right the outputs go to positive infinity; from the left they go to negative infinity.

Guessing game checkpoint

Using this lesson with edumath and SymPy

Use SymPy to check a limit after you have chosen a method by hand.

import sympy as sp

x = sp.symbols("x")
sp.limit((x**2 - 9)/(x - 3), x, 3)

\(\displaystyle 6\)

sp.limit((sp.sqrt(x + 1) - 1) / x, x, 0)

\(\displaystyle \frac{1}{2}\)

sp.limit(1 / x, x, 0, dir="+")
sp.limit(1 / x, x, 0, dir="-")

\(\displaystyle -\infty\)

from edumath.calculus import limit_factor_cancel_steps, validate_limit_answer

solution = limit_factor_cancel_steps("(x**2 - 9)/(x - 3)", 3)
print(solution.answer)
print(validate_limit_answer(6, "(x**2 - 9)/(x - 3)", 3).message)
6
Correct.

Further reading