Direct Substitution
The simplest and most important technique for evaluating limits is direct substitution: plug the value that x is approaching directly into the function. If the result is a finite number with no division by zero, that number is the limit.
For example, lim(x→3) (2x + 1) = 2(3) + 1 = 7.
7This works for all polynomials because they are continuous everywhere. It also works for rational functions, exponentials, logarithms, and trig functions at points where they are continuous.
x = a exactly when the limit equals the function value: lim(x→a) f(x) = f(a).Direct substitution should always be your first attempt. Only when substitution produces an indeterminate form — most commonly 0/0 — do you need algebraic manipulation. Other indeterminate forms include ∞/∞, 0 · ∞, ∞ − ∞, 0⁰, 1^∞, and ∞⁰.
Recognizing that you have an indeterminate form is the critical first step before choosing which algebraic technique to apply next.
Factoring Technique
When direct substitution gives 0/0, the most common fix is factoring. The idea is that both the numerator and denominator contain a factor of (x − a) that equals zero at the limit point, and canceling this common factor eliminates the indeterminate form.
Consider lim(x→2) (x² − 4)/(x − 2). Substituting gives 0/0.
Factor the numerator: (x − 2)(x + 2)/(x − 2).
Cancel (x − 2) to get x + 2.
Now substitute: 2 + 2 = 4.
The limit is 4A more involved example: lim(x→-1) (x³ + 1)/(x + 1).
The numerator factors as a sum of cubes: (x + 1)(x² − x + 1).
Cancel (x + 1) and substitute: 1 + 1 + 1 = 3.
3Patterns to remember
Key factoring patterns to know: difference of squares a² − b² = (a − b)(a + b), sum/difference of cubes, and trinomial factoring. If the numerator does not factor easily, try synthetic division by (x − a) — if substituting x = a gives zero, then (x − a) is guaranteed to be a factor by the Factor Theorem.
This guarantee is why factoring resolves 0/0 forms so reliably.
Rationalizing Technique
When the expression involves a square root and direct substitution gives 0/0, multiply the numerator and denominator by the conjugate to eliminate the radical.
Consider lim(x→0) (√(x + 4) − 2)/x. Substituting gives (2 − 2)/0 = 0/0.
Multiply top and bottom by √(x + 4) + 2: the numerator becomes (x + 4) − 4 = x by the difference-of-squares pattern (a − b)(a + b) = a² − b².
Now you have x / (x(√(x + 4) + 2)). Cancel x: 1/(√(x + 4) + 2).
Substitute x = 0: 1/(2 + 2) = 1/4.
1/4The same technique works when the radical is in the denominator. For lim(x→9) (x − 9)/(√x − 3):
Multiply by (√x + 3)/(√x + 3) to get (x − 9)(√x + 3)/(x − 9) = √x + 3.
Substitute: 3 + 3 = 6.
6The rationalizing technique is also used with cube roots and other radicals, though the conjugate expressions become more complex.
Whenever you see a difference of two terms and one of them contains a root, think "multiply by the conjugate" — it is almost always the right move.
Limits with Trigonometric Functions
lim(x→0) sin(x)/x = 1 and lim(x→0) (1 − cos(x))/x = 0.These cannot be derived by simple algebra; they come from geometric arguments involving the unit circle (or from L'Hopital's rule, which itself relies on derivatives built from these limits). Most trig limit problems reduce to one of these two forms through algebraic manipulation.
For example, lim(x→0) sin(5x)/x:
Multiply and divide by 5 to get 5 · sin(5x)/(5x).
As x → 0, 5x → 0, so sin(5x)/(5x) → 1.
5Another example: lim(x→0) tan(x)/x = lim(x→0) (sin(x)/x) · (1/cos(x)) = 1 · 1 = 1.
1For lim(x→0) sin(3x)/sin(7x):
Rewrite as (sin(3x)/(3x)) · (7x/sin(7x)) · (3x/7x).
Each ratio approaches 1, and the remaining factor is 3/7.
3/7lim(x→0) sin(ax)/sin(bx) = a/b.These tricks appear frequently on AP Calculus and university exams. If a trig limit seems complicated, try rewriting everything in terms of sin(x)/x using identities — Solver AI excels at showing this transformation step by step.
Limits at Infinity
Limits at infinity describe the end behavior of a function — what happens to f(x) as x → ∞ or x → −∞. For rational functions (polynomial divided by polynomial), the technique is to divide every term by the highest power of x in the denominator.
Consider lim(x→∞) (3x² + 2x)/(5x² − 1).
Divide every term by x²: (3 + 2/x)/(5 − 1/x²).
As x → ∞, the terms 2/x and 1/x² approach 0, leaving 3/5.
3/5The degree rule for rational functions
This illustrates the general rule by comparing the degrees of numerator and denominator.
| Case | Condition | Result | Why |
|---|---|---|---|
| 1 | Degrees are equal | Ratio of leading coefficients | Lower-order terms vanish, leaving the leading coefficients. |
| 2 | Numerator's degree < denominator's | 0 | Denominator grows faster than numerator. |
| 3 | Numerator's degree > denominator's | ±∞ | The function grows without bound. |
For functions involving radicals, such as lim(x→∞) (√(4x² + 1) − 2x):
Multiply by the conjugate: (4x² + 1 − 4x²)/(√(4x² + 1) + 2x) = 1/(√(4x² + 1) + 2x).
As x → ∞, the denominator grows without bound, so the limit is 0.
0Exponential functions dominate polynomials at infinity: lim(x→∞) x^n/e^x = 0 for any n, and lim(x→∞) ln(x)/x = 0.
Practice Problems
Problem 1: lim(x→4) (x² − 16)/(x − 4).
Solution: Factor: (x − 4)(x + 4)/(x − 4) = x + 4.
Substitute: 4 + 4 = 8.
8Problem 2: lim(x→0) (√(1 + x) − 1)/x.
Solution: Multiply by conjugate: (1 + x − 1)/(x(√(1 + x) + 1)) = 1/(√(1 + x) + 1).
Substitute: 1/2 = 1/2.
1/2Problem 3: lim(x→0) sin(4x)/(3x).
Solution: Rewrite as (4/3) · sin(4x)/(4x) → (4/3)(1) = 4/3.
4/3Problem 4: lim(x→∞) (7x³ − x)/(2x³ + 5).
Solution: Degrees are equal, so the limit is the ratio of leading coefficients: 7/2.
7/2Problem 5: lim(x→1) (x³ − 1)/(x − 1).
Solution: Factor the difference of cubes: (x − 1)(x² + x + 1)/(x − 1) = x² + x + 1.
Substitute: 1 + 1 + 1 = 3.
3If you want to check your work or explore additional limit techniques like L'Hopital's rule, snap a photo of the problem with Solver AI for an instant step-by-step breakdown.