How to find limits by hand

When a limit problem lands in front of you, the single most important decision is not which formula to use, it is which technique to try first, and understanding how to find limits begins with that choice. Most students lose points not because they cannot do the algebra, but because they reach for the wrong tool and force it. The reliable sequence is short: substitute, then simplify, then escalate. If direct substitution gives you a real number, you are done; the limit is that number. If it gives you 0/0 or ∞/∞, you have an indeterminate form and the real work begins. The order in which to try each method, with one worked example per technique, means you never guess when you sit down to solve a limit.

The question of how to find limits is really a question about method, not memory. You do not need to know twenty tricks; you need to know which one applies to the expression in front of you. Direct substitution is the first move for every limit, because it is fast and because it works whenever the function is continuous at the point. Continuity is not a mystery: a polynomial is continuous everywhere, a rational function is continuous wherever its denominator is not zero, and trig functions are continuous on their domains. If the function is continuous at the point you are approaching, the limit equals the function value, period. That single fact resolves a surprising share of the problems you will meet.

Substitution is not a fallback; it is the default. Write the limit statement, plug the number into the expression, and evaluate. If you get a defined value, that is the answer. No further steps, no algebra, no derivatives. The only reason to do anything else is that substitution failed, and the way it fails tells you what to do next.

Polynomial by Direct Substitution

Evaluate lim_x→3 (2x² − 5x + 1). Substitute x = 3 directly: 2(9) − 15 + 1 = 18 − 15 + 1 = 4. Because 2x² − 5x + 1 is a polynomial, it is continuous everywhere, so the limit as x approaches 3 is exactly the function value at 3, which is 4. That is the entire solution. When substitution gives a nonzero number over zero, something like 5/0, the limit is infinite or does not exist, and you need one-sided limits to decide which. When substitution gives 0/0, the expression is indeterminate and you move to algebra.

The 0/0 form means both numerator and denominator vanish at the point, which usually means they share a factor. Factoring is the fastest way to expose it. Factor the numerator and denominator completely, cancel the common factor, then substitute again. The cancelled factor is never zero at the point in the original function, it is zero only at the point, which is exactly why the original was undefined there, but the simplified function agrees with the original everywhere except at that single point, and limits do not care about the value at the point.

Cancelling a Removable Discontinuity

Evaluate lim_x→2 (x² − 4)/(x − 2). Substitution gives (4 − 4)/(2 − 2) = 0/0, indeterminate. Factor the numerator: x² − 4 = (x − 2)(x + 2). The expression becomes (x − 2)(x + 2)/(x − 2). Cancel the common factor (x − 2), valid for all x ≠ 2, leaving x + 2. Now substitute: 2 + 2 = 4. The limit is 4. The function has a hole at x = 2, but the limit exists and equals 4. This is the classic case where a limit exists even though the function does not, a distinction that separates students who understand limits from those who only compute them.

When the numerator or denominator contains a square root and substitution gives 0/0, multiplying by the conjugate clears the radical. The conjugate of a + b is a − b; multiplying them uses the difference of squares to remove the root. This is a limit technique that appears constantly in AP Calculus AB and BC, and it is the one students most often skip because they do not recognize the pattern. The trigger is simple: a radical plus or minus a constant, with zero in the denominator after substitution.

Rationalising the Numerator

Evaluate lim_x→4 (√x − 2)/(x − 4). Substitution gives (2 − 2)/(4 − 4) = 0/0. Multiply numerator and denominator by the conjugate of the numerator, √x + 2: [(√x − 2)(√x + 2)] / [(x − 4)(√x + 2)]. The numerator becomes x − 4. The expression is now (x − 4)/[(x − 4)(√x + 2)]. Cancel (x − 4), leaving 1/(√x + 2). Substitute x = 4: 1/(2 + 2) = 1/4. The limit is 1/4. Note that you multiply by the conjugate of whichever part contains the radical, here the numerator, and that you never multiply out the other factor, because you need it to cancel.

Complex fractions, a fraction inside a fraction, produce 0/0 when the inner terms cancel at the point. The fix is to combine the numerator or denominator into a single fraction first, then simplify. This technique is really factoring in disguise, but the factoring happens after you clear the nested denominators. It is the go-to move for limits involving sums of rational expressions, and it is a solving limit step that many textbooks bury inside other problems.

Clearing a Complex Fraction

Evaluate lim_x→0 [(1/(x+3)) − (1/3)] / x. Substitution gives [(1/3) − (1/3)] / 0 = 0/0. Combine the numerator over a common denominator: [3 − (x + 3)] / [3(x + 3)] = (−x) / [3(x + 3)]. The full expression is now [−x / (3(x + 3))] / x = −x / [3x(x + 3)]. Cancel x, leaving −1 / [3(x + 3)]. Substitute x = 0: −1 / [3(3)] = −1/9. The limit is −1/9. The key is recognising that the numerator is a difference of fractions, not a single term, and that combining them is the only way to expose the factor that cancels.

L'Hôpital's rule is the heavy artillery, and like all heavy artillery it is precise about when it fires. It applies only to indeterminate forms of type 0/0 or ∞/∞. You differentiate the numerator and the denominator separately, then take the limit again. If the result is still 0/0 or ∞/∞, you apply it again, there is no limit to how many times, as long as each application keeps the form indeterminate. What it is not for: forms like 0·∞, 1^∞, ∞ − ∞, 0^0, or ∞^0. Those must be rewritten algebraically into 0/0 or ∞/∞ before the rule applies, and misapplying it to a non-indeterminate form is the single most common error in this subject.

Repeated Application

Evaluate lim_x→0 (sin x − x)/x³. Substitution gives (0 − 0)/0 = 0/0. Apply L'Hôpital's rule: differentiate numerator and denominator. d/dx(sin x − x) = cos x − 1, and d/dx(x³) = 3x². The new limit is (cos x − 1)/(3x²), still 0/0. Apply the rule again: d/dx(cos x − 1) = −sin x, d/dx(3x²) = 6x. Now the limit is (−sin x)/(6x), still 0/0. Apply it a third time: d/dx(−sin x) = −cos x, d/dx(6x) = 6. The limit is (−cos 0)/6 = −1/6. The rule worked three times in a row, and each application was justified because the form remained indeterminate throughout.

The squeeze theorem is the only technique on this list that does not compute a limit directly, it proves one exists by trapping the function between two others that share the same limit. It is the right tool when the function oscillates, typically involving sin(1/x), cos(1/x), or a product of a bounded function with one that approaches zero. The classic setup: you know −1 ≤ sin(1/x) ≤ 1, so you multiply through by something like x², and both the lower and upper bounds approach the same number.

Bounding an Oscillating Function

Evaluate lim_x→0 x² sin(1/x). The sine term oscillates wildly as x approaches 0, it has no limit, but it is bounded: −1 ≤ sin(1/x) ≤ 1 for all x ≠ 0. Multiply through by x², which is nonnegative: −x² ≤ x² sin(1/x) ≤ x². Now take limits as x→0 on the outer two functions. lim_x→0 (−x²) = 0 and lim_x→0 x² = 0. Since x² sin(1/x) is squeezed between two functions that both approach 0, the squeeze theorem forces lim_x→0 x² sin(1/x) = 0. The limit exists and equals 0, even though the function itself never settles down near the point. That is the entire point of the theorem: it converts a statement about oscillation into a statement about bounds.

Here is the order of operations that resolves nearly every manual limit problem. First, try direct substitution. If you get a real number, stop. If you get 0/0 or ∞/∞, ask what kind of expression you have. If it is a rational function, factor and cancel. If it has a radical, multiply by the conjugate. If it is a complex fraction, combine the fractions. If it is a trig function involving sin or cos of a rational expression that oscillates, use the squeeze theorem. If you have already simplified and still have 0/0 or ∞/∞, then, and only then, apply L'Hôpital's rule. If the function is piecewise or involves absolute value, check the one-sided limits separately before deciding anything. That last step is the one most students skip, and it is the reason they write DNE when the two-sided limit actually exists.

Substitution Gives	Expression Type	Technique to Try	Example Result
Real number	Any continuous function	Direct substitution	limx→3(2x²−5x+1) = 4
0/0	Rational function	Factor and cancel	limx→2(x²−4)/(x−2) = 4
0/0	Radical in numerator/denominator	Multiply by conjugate	limx→4(√x−2)/(x−4) = 1/4
0/0	Complex fraction	Combine fractions	limx→0[(1/(x+3))−(1/3)]/x = −1/9
0/0 or ∞/∞	Any differentiable pair	L'Hôpital's rule	limx→0(sin x−x)/x³ = −1/6
Oscillation	Bounded × →0	Squeeze theorem	limx→0 x² sin(1/x) = 0

The most common failure is applying direct substitution to a function that is not continuous at the point, then concluding the limit does not exist when substitution gives a nonzero number over zero. That result means the limit is infinite or DNE, but it does not tell you which, you must check the one-sided limits. If the function approaches +∞ from one side and −∞ from the other, the two-sided limit is DNE. If both sides go to +∞, the limit is ∞, which is a specific kind of DNE, the function grows without bound, and Stewart §2.2 allows writing lim = ∞, though it is not a real number.

Another failure is applying L'Hôpital's rule to a form that is not 0/0 or ∞/∞. The rule is silent on 0·∞, and applying it there gives nonsense. Rewrite 0·∞ as a quotient first, put one factor in the denominator, then recheck the form. A third failure is the squeeze theorem with incorrect bounds: if your lower and upper functions do not share the same limit, the theorem does not apply, and you have proven nothing. Finally, do not assume a limit exists from a table of values. A table only samples finitely many points, and a function can jump between them. The limit must be forced by the definition, not guessed from data.