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Evaluating Limits — Definition, Formula & Examples

Evaluating limits is the process of finding the value that a function approaches as its input gets closer and closer to a specific number. You can evaluate limits using direct substitution, algebraic manipulation, or special techniques like L'Hôpital's Rule.

To evaluate lim⁡x→cf(x)\lim_{x \to c} f(x) is to determine the real number LL (if it exists) such that for every ε>0\varepsilon > 0 there exists a δ>0\delta > 0 where 0<∣x−c∣<δ0 < |x - c| < \delta implies ∣f(x)−L∣<ε|f(x) - L| < \varepsilon. In practice, evaluation proceeds by substitution when ff is continuous at cc, or by algebraic simplification and limit laws when substitution yields an indeterminate form.

Key Formula

lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L
Where:
  • xx = The input variable approaching the target value
  • cc = The value that x approaches
  • f(x)f(x) = The function being evaluated
  • LL = The limit value the function approaches

How It Works

Start by plugging the target value directly into the function. If you get a real number, that number is the limit. If substitution produces an indeterminate form like 00\frac{0}{0}, you need to simplify the expression first — typically by factoring, canceling common factors, or rationalizing a numerator or denominator. Once the problematic term is eliminated, substitute again. For limits at infinity, divide every term by the highest power of xx in the denominator and observe which terms vanish.

Worked Example

Problem: Evaluate lim⁡x→3x2−9x−3\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}.
Step 1: Try direct substitution: Substitute x=3x = 3 into the expression.
32−93−3=00\frac{3^2 - 9}{3 - 3} = \frac{0}{0}
Step 2: Factor the numerator: Recognize x2−9x^2 - 9 as a difference of squares.
x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3)
Step 3: Cancel the common factor: Cancel the (x−3)(x - 3) that appears in both the numerator and denominator. This is valid because x≠3x \neq 3 as xx approaches 3.
(x−3)(x+3)x−3=x+3\frac{(x - 3)(x + 3)}{x - 3} = x + 3
Step 4: Substitute again: Now plug in x=3x = 3 to the simplified expression.
3+3=63 + 3 = 6
Answer: lim⁡x→3x2−9x−3=6\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} = 6

Another Example

Problem: Evaluate lim⁡x→0x+4−2x\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x}.
Step 1: Try direct substitution: Substitute x=0x = 0.
0+4−20=2−20=00\frac{\sqrt{0 + 4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}
Step 2: Rationalize the numerator: Multiply the numerator and denominator by the conjugate x+4+2\sqrt{x + 4} + 2.
x+4−2x⋅x+4+2x+4+2=(x+4)−4x(x+4+2)=xx(x+4+2)\frac{\sqrt{x+4}-2}{x} \cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \frac{(x+4)-4}{x(\sqrt{x+4}+2)} = \frac{x}{x(\sqrt{x+4}+2)}
Step 3: Cancel and substitute: Cancel the xx terms, then substitute x=0x = 0.
1x+4+2  ∣x=0=14+2=14\frac{1}{\sqrt{x+4}+2} \;\bigg|_{x=0} = \frac{1}{\sqrt{4}+2} = \frac{1}{4}
Answer: lim⁡x→0x+4−2x=14\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x} = \dfrac{1}{4}

Why It Matters

Evaluating limits is the gateway skill for all of AP Calculus AB/BC. Derivatives are defined as limits, and integrals rely on limit processes, so nearly every later topic depends on this one. Engineers and physicists use limits constantly — for instance, when modeling instantaneous velocity or the behavior of circuits at boundary conditions.

Common Mistakes

Mistake: Concluding the limit does not exist just because substitution gives 00\frac{0}{0}.
Correction: The form 00\frac{0}{0} is indeterminate, not undefined. It signals that you must simplify before evaluating. The limit often does exist after canceling or rationalizing.
Mistake: Canceling a factor like (x−3)(x - 3) without recognizing x≠3x \neq 3.
Correction: You can cancel because limits examine values near cc, not at cc itself. But remember: the simplified function equals the original everywhere except possibly at x=cx = c. The function value at cc may still differ from the limit.

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