Finding Limits Algebraically When Direct Substitution Fails

For many functions, evaluating a limit is as simple as plugging the target value straight into the function. This is called the direct substitution property, and it works whenever the function is defined and continuous at that point. But sometimes substitution gives you something meaningless, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These are called indeterminate forms, and they do not tell you the value of the limit at all — they tell you that you need a different strategy.

The good news is that an indeterminate form almost always signals that the expression can be rewritten. Once it is rewritten, direct substitution works again on the new, simplified expression.

Suppose you want \(\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}\). Substituting \(x = 2\) gives \(\dfrac{4 - 4}{2 - 2} = \dfrac{0}{0}\). That is not a real number, so substitution alone cannot answer the question. Notice, though, that the numerator factors:

\(\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2}\)

For every \(x \ne 2\), the \((x-2)\) factors cancel, leaving \(x+2\). Since a limit only cares about values near \(x=2\), not at \(x=2\) itself, this cancellation is completely valid:

\(\lim_{x \to 2} \dfrac{x^2-4}{x-2} = \lim_{x \to 2} (x+2) = 4\)

Graphically, the original function has a hole at \(x = 2\), but the limit still exists there and equals 4.

This is the go-to first move whenever you see \(\frac{0}{0}\) with polynomials. Factor the numerator, the denominator, or both, then cancel any factor that is common to both and causes the zero. Common patterns to look for include difference of squares, \(a^2 - b^2 = (a-b)(a+b)\), perfect square trinomials, and simple grouping.

Example. Find \(\lim_{x \to 3} \dfrac{x^2 - x - 6}{x - 3}\).

Substitution gives \(\frac{0}{0}\). Factor the numerator: \(x^2 - x - 6 = (x-3)(x+2)\). So

\(\lim_{x \to 3} \dfrac{(x-3)(x+2)}{x-3} = \lim_{x \to 3} (x+2) = 5\)

When a square root creates the indeterminate form, multiply the numerator and denominator by the conjugate of the radical expression. This clears the root from one part of the fraction so a common factor can cancel.

Example. Find \(\lim_{x \to 0} \dfrac{\sqrt{x+1} - 1}{x}\).

Direct substitution gives \(\frac{0}{0}\). Multiply top and bottom by the conjugate \(\sqrt{x+1}+1\):

\(\dfrac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{x(\sqrt{x+1}+1)} = \dfrac{(x+1) - 1}{x(\sqrt{x+1}+1)} = \dfrac{x}{x(\sqrt{x+1}+1)}\)

Cancelling the \(x\) leaves \(\dfrac{1}{\sqrt{x+1}+1}\), and now substitution works: \(\dfrac{1}{\sqrt{0+1}+1} = \dfrac{1}{2}\).

When the indeterminate form comes from a difference of fractions, such as \(\dfrac{1}{x+h} - \dfrac{1}{x}\) type expressions common in derivative definitions, first combine everything over one common denominator, simplify the numerator, and then cancel.

Example. Find \(\lim_{h \to 0} \dfrac{\frac{1}{2+h} - \frac{1}{2}}{h}\).

Combine the top fraction: \(\dfrac{1}{2+h} - \dfrac{1}{2} = \dfrac{2 - (2+h)}{2(2+h)} = \dfrac{-h}{2(2+h)}\). Dividing by \(h\) gives \(\dfrac{-1}{2(2+h)}\), and substituting \(h = 0\) now gives \(-\dfrac{1}{4}\).

The same techniques apply when a limit is restricted to one side, written \(\lim_{x \to a^-}\) (from the left) or \(\lim_{x \to a^+}\) (from the right). For piecewise functions, substitute using whichever piece is valid on that side of \(a\). For rational functions with a factor that changes sign at \(a\), factor and cancel first, then check the sign of the remaining expression as \(x\) approaches \(a\) from the chosen side. One-sided reasoning is also essential when a function has a vertical asymptote rather than a removable hole; that case is covered in more depth on the infinite limits and vertical asymptotes page.

A limit that exists algebraically at a point does not automatically mean the function is continuous there — in the hole example above, the limit is 4, but the function itself is undefined at \(x=2\). Comparing a limit's value to the function's actual output at that point is exactly the idea behind continuity, so mastering algebraic limit techniques here sets you up directly for that topic.

When direct substitution fails, work through these steps:

1. Confirm the result really is indeterminate, such as \(\frac{0}{0}\), rather than a genuine undefined case like \(\frac{5}{0}\) (which signals a vertical asymptote instead).
2. Look for a factorable polynomial and cancel common factors.
3. If a square root is present, multiply by the conjugate.
4. If there are fractions within the expression, combine them into a single fraction first.
5. Re-substitute into the simplified expression to find the limit.

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