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Evaluating Limits






When Substitution Fails


The simplest approach to any limit is direct substitution: plug in the value and compute. For polynomials, this always works. For rational functions away from zeros of the denominator, it works just as well. But substitution has limits of its own.

When plugging in produces 0/00/0, the expression is indeterminate—neither the numerator nor denominator alone determines the result. The limit might be any finite number, or infinite, or nonexistent. The form 0/00/0 signals that cancellation is hiding the true behavior, and algebraic work is required to reveal it.

This page covers the core techniques: factoring, rationalizing, and algebraic manipulation. Each method transforms an indeterminate expression into one where substitution succeeds.



Direct Substitution — Try This First


    Before attempting any technique, substitute aa into f(x)f(x) directly. If the result is a finite number with no division by zero, that number is the limit:

    limxaf(x)=f(a)\lim_{x \to a} f(x) = f(a)


    This works for:

  • aa

  • Direct substitution exploits continuity. When ff is continuous at aa, the limit equals the function value by definition.

When Direct Substitution Fails


Substitution fails when it produces an undefined or indeterminate expression. The most common outcome is 0/00/0: both numerator and denominator evaluate to zero.

For example:

limx2x24x2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}


Substituting x=2x = 2 gives 00\dfrac{0}{0}. This does not mean the limit is zero, undefined, or nonexistent. It means the expression's behavior near x=2x = 2 is not yet determined—more work is needed.

The 0/00/0 form indicates that both numerator and denominator share a common factor of (x2)(x - 2). Removing this factor reveals the limit.

Indeterminate Forms


Several forms signal that limit rules cannot be applied directly:

000\frac{0}{0} \qquad \frac{\infty}{\infty} \qquad 0 \cdot \infty \qquad \infty - \infty


00100^0 \qquad 1^\infty \qquad \infty^0


Each form represents a competition between opposing tendencies. In 0/00/0, both numerator and denominator vanish—which vanishes faster determines the limit. In \infty - \infty, both terms grow without bound—their difference depends on relative growth rates.

Indeterminate forms require transformation. The goal is to rewrite the expression so that substitution or limit rules apply.

Factoring and Canceling


When substitution yields 0/00/0, the numerator and denominator share a common factor. Factor both, cancel the shared factor, then substitute.

limx2x24x2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}


Factor the numerator:

=limx2(x2)(x+2)x2= \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2}


Cancel (x2)(x - 2):

=limx2(x+2)=4= \lim_{x \to 2} (x + 2) = 4


The cancellation is valid because the limit considers xx near 22, not at 22. For x2x \neq 2, the factor (x2)(x - 2) is nonzero and cancels legitimately.

Rationalizing — Conjugate Multiplication


When radicals appear and substitution fails, multiply by the conjugate to eliminate the radical.

limx0x+11x\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}


Substituting gives 0/00/0. Multiply numerator and denominator by x+1+1\sqrt{x + 1} + 1:

=limx0(x+11)(x+1+1)x(x+1+1)= \lim_{x \to 0} \frac{(\sqrt{x + 1} - 1)(\sqrt{x + 1} + 1)}{x(\sqrt{x + 1} + 1)}


The numerator becomes a difference of squares:

=limx0(x+1)1x(x+1+1)=limx0xx(x+1+1)= \lim_{x \to 0} \frac{(x + 1) - 1}{x(\sqrt{x + 1} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 1} + 1)}


Cancel xx:

=limx01x+1+1=12= \lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1} = \frac{1}{2}


Expanding and Simplifying


Sometimes expanding a product or simplifying a complex fraction reveals the cancellation needed.

limx1(x+1)24x1\lim_{x \to 1} \frac{(x + 1)^2 - 4}{x - 1}


Expand the numerator:

=limx1x2+2x+14x1=limx1x2+2x3x1= \lim_{x \to 1} \frac{x^2 + 2x + 1 - 4}{x - 1} = \lim_{x \to 1} \frac{x^2 + 2x - 3}{x - 1}


Factor:

=limx1(x1)(x+3)x1=limx1(x+3)=4= \lim_{x \to 1} \frac{(x - 1)(x + 3)}{x - 1} = \lim_{x \to 1} (x + 3) = 4


The initial form obscured the factor of (x1)(x - 1); expansion made it visible.

Combining Fractions


When the expression involves a difference of fractions, combine them over a common denominator.

limx1(1x12x21)\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)


Note that x21=(x1)(x+1)x^2 - 1 = (x - 1)(x + 1). Rewrite with common denominator:

=limx1(x+1(x1)(x+1)2(x1)(x+1))= \lim_{x \to 1} \left( \frac{x + 1}{(x - 1)(x + 1)} - \frac{2}{(x - 1)(x + 1)} \right)


=limx1x+12(x1)(x+1)=limx1x1(x1)(x+1)= \lim_{x \to 1} \frac{x + 1 - 2}{(x - 1)(x + 1)} = \lim_{x \to 1} \frac{x - 1}{(x - 1)(x + 1)}


Cancel (x1)(x - 1):

=limx11x+1=12= \lim_{x \to 1} \frac{1}{x + 1} = \frac{1}{2}


Multiplying by Strategic Forms of 1


For limits at infinity, divide numerator and denominator by the highest power of xx in the denominator.

limx3x2+5x12x27\lim_{x \to \infty} \frac{3x^2 + 5x - 1}{2x^2 - 7}


Divide every term by x2x^2:

=limx3+5x1x227x2= \lim_{x \to \infty} \frac{3 + \frac{5}{x} - \frac{1}{x^2}}{2 - \frac{7}{x^2}}


As xx \to \infty, terms with xx in the denominator vanish:

=3+0020=32= \frac{3 + 0 - 0}{2 - 0} = \frac{3}{2}


This technique isolates the dominant terms that control behavior at infinity.

Using Known Limits


Recognize when parts of an expression match special limits and rewrite accordingly.

limx0sin3xx\lim_{x \to 0} \frac{\sin 3x}{x}


Rewrite to match the standard form sinuu\dfrac{\sin u}{u}:

=limx0sin3xx33=3limx0sin3x3x= \lim_{x \to 0} \frac{\sin 3x}{x} \cdot \frac{3}{3} = 3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x}


Let u=3xu = 3x. As x0x \to 0, u0u \to 0:

=3limu0sinuu=31=3= 3 \cdot \lim_{u \to 0} \frac{\sin u}{u} = 3 \cdot 1 = 3


One-Sided Evaluation


When one-sided limits differ, evaluate each separately.

limx0xx\lim_{x \to 0} \frac{|x|}{x}


For x>0x > 0: x=x|x| = x, so xx=1\dfrac{|x|}{x} = 1

For x<0x < 0: x=x|x| = -x, so xx=1\dfrac{|x|}{x} = -1

limx0+xx=1limx0xx=1\lim_{x \to 0^+} \frac{|x|}{x} = 1 \qquad \lim_{x \to 0^-} \frac{|x|}{x} = -1


The one-sided limits differ, so the two-sided limit does not exist.

Sign Analysis Near the Point


When a limit involves potential division by zero with a nonzero numerator, determine the sign to identify ++\infty or -\infty.

limx3+x+1x3\lim_{x \to 3^+} \frac{x + 1}{x - 3}


At x=3x = 3: numerator =4>0= 4 > 0, denominator 0\to 0.

For xx slightly greater than 33: x3>0x - 3 > 0 (small positive).

Positive divided by small positive gives large positive:

limx3+x+1x3=+\lim_{x \to 3^+} \frac{x + 1}{x - 3} = +\infty


From the left, x3<0x - 3 < 0, so:

limx3x+1x3=\lim_{x \to 3^-} \frac{x + 1}{x - 3} = -\infty


Worked Examples


Example 1: Factoring


limx3x29x3=limx3(x3)(x+3)x3=limx3(x+3)=6\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3}(x + 3) = 6


Example 2: Rationalizing


limx4x2x4=limx4(x2)(x+2)(x4)(x+2)=limx4x4(x4)(x+2)=14\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \lim_{x \to 4} \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)} = \lim_{x \to 4} \frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{4}


Example 3: Using Special Limits


limx01cosxx2=12\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}


This follows from the special limit limx01cosxx2=12\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}.

Example 4: Combining Fractions


limx2(1x24x24)=limx2x+24(x2)(x+2)=limx2x2(x2)(x+2)=14\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^2 - 4} \right) = \lim_{x \to 2} \frac{x + 2 - 4}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x - 2}{(x-2)(x+2)} = \frac{1}{4}