Evaluating Limits Algebraically: Methods and Examples

What Does Evaluating Limits Algebraically Mean?

Evaluating limits algebraically means finding $\lim_{x \to a} f(x)$ by manipulating the formula for $f(x)$, rather than by reading a graph or building a table of values. The method always starts the same way: substitute $x = a$. If that gives a real number, you are done. If it gives $\frac{0}{0}$, you rewrite the expression (factor, rationalize, combine fractions, or divide by a power of $x$) and substitute again.

Recall what the limit itself says. The notation $\lim_{x \to a} f(x) = L$ means that the outputs $f(x)$ get as close as you like to $L$ when $x$ is close enough to $a$, with $x \neq a$. A full treatment of that definition lives on the limit of a function page. For this page, one consequence matters more than any other:

$$\text{If } f(x) = g(x) \text{ for every } x \neq a \text{ near } a, \text{ then } \lim_{x \to a} f(x) = \lim_{x \to a} g(x).$$

• $f$ is the original expression, often undefined at $a$.

• $g$ is the rewritten expression, which agrees with $f$ everywhere near $a$ except possibly at $a$ itself.

• The condition "for every $x \neq a$ near $a$" is the whole licence. The value at $a$ never enters a limit.

The arithmetic you use along the way (the limit of a sum is the sum of the limits, and so on) comes from the limit laws. This page assumes those laws and focuses on the rewrites.

Why Can You Cancel A Factor When Direct Substitution Gives 0/0?

You can cancel because a limit never looks at $x = a$ itself, and at every other input the cancelled factor is a nonzero number divided by itself. Two Physics Forums threads ask this exact question, "Why does cancellation in limits work like this?" and "Justification for cancelling terms in limits?", and the worry behind both is fair: cancelling $(x-3)$ when $x = 3$ looks like dividing by zero.

Take $f(x) = \frac{x^2 - 9}{x - 3}$. For any $x \neq 3$, the factor $x - 3$ is not zero, so

$$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3 \quad (x \neq 3).$$

The two functions $f(x)$ and $g(x) = x + 3$ are different functions. Their domains differ, since $f$ is undefined at $3$ and $g$ is not. But they agree at every input except $3$, and the limit as $x \to 3$ only uses inputs near $3$, never $3$ itself. So the limit of $f$ equals the limit of $g$, which is $6$ by plain substitution.

Geometrically, the graph of $f$ is the line $y = x + 3$ with a single point punched out at $(3, 6)$. The algebra fills that hole, and the filled-in height is the limit. A hole like this is a removable discontinuity, and evaluating limits algebraically is how you find where the missing point belongs.

Which Technique Should You Try First?

Substitute first, then let the shape of the expression choose the technique. A Quora question captures the confusion well: "If I am doing limit calculus and the answer I receive is 0/0, when do I use conjugates? When do I claim that the limit does not exist?" The short answer is that $\frac{0}{0}$ never means "does not exist". It means "rewrite and try again".

Run the table top to bottom and stop at the first row that fits. Two cases fall outside it: a bounded, oscillating factor such as $x^2 \sin\frac{1}{x}$, which needs the squeeze theorem, and a $\frac{0}{0}$ quotient that resists every rewrite, where L'Hôpital's rule (differentiating top and bottom separately) is the usual next tool.

How Does Direct Substitution Work?

Direct substitution works whenever the function is continuous at $a$: then the limit equals the function value, $\lim_{x \to a} f(x) = f(a)$. Polynomials are continuous everywhere, and a quotient of polynomials (a rational function) is continuous wherever its denominator is not zero, so for these you simply plug in. The idea that "the value and the limit agree" is exactly what continuity of a function means.

$$\lim_{x \to 2} \left(3x^2 - 5x + 1\right) = 3(4) - 10 + 1 = 3$$

$$\lim_{x \to 1} \frac{x + 4}{x^2 + 1} = \frac{5}{2}$$

The habit to build is to substitute before you do anything clever. Students who reach for factoring by reflex sometimes spend a page simplifying a limit that substitution answers in one line.

How Do You Evaluate A Limit By Factoring?

When substitution gives $\frac{0}{0}$ and both numerator and denominator are polynomials, $x - a$ is a factor of both (by the factor theorem), so factor it out and cancel the common factor. Then substitute again.

$$\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2}$$

$$= \lim_{x \to 2} \frac{(x - 2)(x + 3)}{x - 2}$$

$$= \lim_{x \to 2} (x + 3)$$

$$= 5$$

The factoring patterns you need are the ones from algebra: common factors, trinomials, the difference of squares $a^2 - b^2 = (a - b)(a + b)$, and the difference of cubes $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. When no pattern is obvious, divide the polynomial by $x - a$ directly. Since $x = a$ makes the numerator zero, the division always comes out exact.

Notice the "lim" stays on every line until the very last one. Writing $\frac{(x-2)(x+3)}{x-2} = x + 3$ without the limit (or without the condition $x \neq 2$) is a false statement at $x = 2$.

How Do You Rationalize To Evaluate A Limit?

When a square root sits in a difference such as $\sqrt{x} - 3$, multiply the numerator and the denominator by the conjugate $\sqrt{x} + 3$. The product $(\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9$ has no root left, and the factor that caused the zero becomes visible.

$$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$$

$$= \lim_{x \to 9} \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)}$$

$$= \lim_{x \to 9} \frac{x - 9}{(x - 9)(\sqrt{x} + 3)}$$

$$= \lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{6}$$

A tutor on Physics Forums ("The Conjugate Method for Limits") tried this on $\frac{f(x)}{x - 3}$ by multiplying by $x + 3$ and found the denominator still went to zero. That is the right observation.

The conjugate method only helps when the zero comes from a difference of square roots: the conjugate turns that difference into a polynomial that shares a factor with the other side. Multiplying a plain $x - 3$ by $x + 3$ creates no common factor, so nothing cancels. Use the conjugate on the part that contains the root, whether it sits on top or on the bottom.

How Do You Simplify Complex Fractions In A Limit?

When the expression has fractions inside a fraction, combine the small fractions over a common denominator first. The factor $x - a$ usually appears in the new numerator and cancels.

$$\lim_{x \to 0} \frac{\frac{1}{x + 2} - \frac{1}{2}}{x}$$

$$= \lim_{x \to 0} \frac{\frac{2 - (x + 2)}{2(x + 2)}}{x}$$

$$= \lim_{x \to 0} \frac{-x}{2x(x + 2)}$$

$$= \lim_{x \to 0} \frac{-1}{2(x + 2)} = -\frac{1}{4}$$

The tempting shortcut is to split the fraction into $\frac{1}{x(x+2)} - \frac{1}{2x}$ and take each limit separately. Each piece blows up as $x \to 0$, which hands you $\infty - \infty$, another indeterminate form. Combining first avoids the trap entirely.

How Do You Evaluate Limits At Infinity By Dividing By The Highest Power?

As $x \to \infty$, divide every term in the numerator and denominator by the highest power of $x$ in the denominator. Every term of the form $\frac{c}{x^n}$ (with $n > 0$) then goes to $0$, and what remains is a plain number.

$$\lim_{x \to \infty} \frac{2x^2 - 3x + 1}{5x^2 + 4}$$

$$= \lim_{x \to \infty} \frac{2 - \frac{3}{x} + \frac{1}{x^2}}{5 + \frac{4}{x^2}}$$

$$= \frac{2 - 0 + 0}{5 + 0} = \frac{2}{5}$$

The same move gives the three standard outcomes for a rational function, which the limits at infinity page develops in detail:

• Same degree on top and bottom: the limit is the ratio of the leading coefficients, as above.

• Lower degree on top: the limit is $0$, for example $\lim_{x \to \infty} \frac{3x + 1}{x^2 + 5} = 0$.

• Higher degree on top: the expression grows without bound, for example $\frac{x^3}{x^2 + 1} \to \infty$.

Square roots need one extra line of care. For $x < 0$, $\sqrt{x^2} = |x| = -x$, not $x$. That sign is the whole difference in Example 5 below.

How Do You Evaluate Trigonometric Limits Algebraically?

Trigonometric limits are evaluated by rewriting the expression so it contains $\frac{\sin u}{u}$ with $u \to 0$, then using the two standard results below. They are proved geometrically with the squeeze theorem, and the standard limits page collects the rest of the family.

$$\lim_{u \to 0} \frac{\sin u}{u} = 1 \qquad \lim_{u \to 0} \frac{1 - \cos u}{u} = 0$$

Both require $u$ in radians. The usual move is to multiply and divide by whatever makes the argument and the denominator match:

$$\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin 5x}{5x} = 5 \cdot 1 = 5$$

For a ratio of two sines, create two copies of the standard limit:

$$\lim_{x \to 0} \frac{\sin 5x}{\sin 2x} = \lim_{x \to 0} \frac{\frac{\sin 5x}{5x} \cdot 5x}{\frac{\sin 2x}{2x} \cdot 2x} = \frac{5}{2}$$

And $\tan x = \frac{\sin x}{\cos x}$ gives $\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1$.

How Do You Find One-Sided Limits Algebraically?

You find a one-sided limit algebraically by replacing the absolute value or the piecewise rule with the formula that holds on that side, then substituting. Quora users ask "How do I find one-sided limits algebraically?", and the answer is that the side you approach from tells you which formula to use.

For $\frac{|x - 2|}{x - 2}$, the absolute value function splits into two cases. When $x > 2$, $|x - 2| = x - 2$, and when $x < 2$, $|x - 2| = -(x - 2)$:

$$\lim_{x \to 2^{+}} \frac{|x - 2|}{x - 2} = \lim_{x \to 2^{+}} \frac{x - 2}{x - 2} = 1$$

$$\lim_{x \to 2^{-}} \frac{|x - 2|}{x - 2} = \lim_{x \to 2^{-}} \frac{-(x - 2)}{x - 2} = -1$$

The one-sided limits disagree, so the two-sided limit does not exist. Piecewise functions work the same way. For $f(x) = x^2 + 1$ when $x < 1$ and $f(x) = 3x - 1$ when $x \geq 1$, the left formula gives $2$ and the right formula gives $2$, so $\lim_{x \to 1} f(x) = 2$.

When Does Algebra Show That A Limit Does Not Exist?

Algebra shows a limit does not exist when substitution gives $\frac{k}{0}$ with $k \neq 0$ and the two sides head in different directions, or when the one-sided limits disagree. The key distinction is between $\frac{0}{0}$, which is one of the indeterminate forms and tells you nothing yet, and $\frac{k}{0}$, which is not indeterminate at all.

With $\frac{k}{0}$, the size of the expression grows without bound, so only the sign is left to decide:

• $\lim_{x \to 1} \frac{1}{(x - 1)^2} = \infty$, because the denominator is positive on both sides.

• $\lim_{x \to 1} \frac{x + 1}{x - 1}$ does not exist: the right-hand limit is $\infty$ and the left-hand limit is $-\infty$.

So the honest workflow is: $\frac{0}{0}$ means rewrite, $\frac{k}{0}$ means check signs, and only disagreeing sides (or unbounded behaviour) justify "does not exist".

Examples Of Evaluating Limits Algebraically

Example 1

Evaluate $\lim_{x \to 3} \dfrac{x^2 + 2x - 1}{x + 1}$.

Substitute $x = 3$. The denominator is $4$, not zero, so the function is continuous at $3$.

$$\frac{9 + 6 - 1}{3 + 1} = \frac{14}{4} = \frac{7}{2}$$

Final answer: $\frac{7}{2}$

Example 2

Evaluate $\lim_{x \to 2} \dfrac{x^3 - 8}{x - 2}$.

Substitution gives $\frac{8 - 8}{2 - 2} = \frac{0}{0}$. A tempting conclusion is that "anything over itself is $1$", so the limit is $1$.

Check that against nearby inputs:

The values settle near $12$, nowhere near $1$. The zero on top and the zero on the bottom shrink at different rates, and $\frac{0}{0}$ carries no information about that ratio. Factor instead, using the difference of cubes:

$$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2}$$

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

$$= 4 + 4 + 4 = 12$$

Final answer: $12$

Example 3

Evaluate $\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x}$.

Substitution gives $\frac{0}{0}$, and the zero comes from a root difference, so multiply by the conjugate $\sqrt{x + 4} + 2$.

$$\lim_{x \to 0} \frac{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)}{x(\sqrt{x + 4} + 2)}$$

$$= \lim_{x \to 0} \frac{(x + 4) - 4}{x(\sqrt{x + 4} + 2)}$$

$$= \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2}$$

$$= \frac{1}{2 + 2} = \frac{1}{4}$$

Check: at $x = 0.001$ the original expression is about $0.24998$.

Final answer: $\frac{1}{4}$

Example 4

Evaluate $\lim_{x \to 3} \dfrac{\frac{1}{x} - \frac{1}{3}}{x - 3}$.

Combine the top over the common denominator $3x$.

$$\frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x} = \frac{-(x - 3)}{3x}$$

$$\lim_{x \to 3} \frac{-(x - 3)}{3x(x - 3)}$$

$$= \lim_{x \to 3} \frac{-1}{3x}$$

$$= -\frac{1}{9}$$

This limit is the slope of the tangent to $y = \frac{1}{x}$ at $x = 3$, a first glimpse of a derivative.

Final answer: $-\frac{1}{9}$

Example 5

Evaluate $\lim_{x \to -\infty} \dfrac{3x}{\sqrt{x^2 + 1}}$.

Divide top and bottom by $|x|$, the size of the highest power under the root. For $x < 0$, $|x| = -x$ and $\sqrt{x^2} = -x$.

$$\frac{3x}{\sqrt{x^2 + 1}} = \frac{\frac{3x}{-x}}{\frac{\sqrt{x^2 + 1}}{\sqrt{x^2}}} = \frac{-3}{\sqrt{1 + \frac{1}{x^2}}}$$

$$\lim_{x \to -\infty} \frac{-3}{\sqrt{1 + \frac{1}{x^2}}} = \frac{-3}{\sqrt{1 + 0}} = -3$$

The sign makes sense: the numerator is negative and the denominator is positive, so the quotient must be negative.

Final answer: $-3$

Example 6

Evaluate $\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}$.

Substitution gives $\frac{0}{0}$. Multiply by the conjugate $1 + \cos x$ and use $1 - \cos^2 x = \sin^2 x$.

$$\lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x)}{x^2(1 + \cos x)}$$

$$= \lim_{x \to 0} \frac{\sin^2 x}{x^2(1 + \cos x)}$$

$$= \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x}$$

$$= 1^2 \cdot \frac{1}{2} = \frac{1}{2}$$

Final answer: $\frac{1}{2}$

Why Did Mathematicians Need A Rule For Cancelling?

Mathematicians needed it because early calculus cancelled a quantity and then set it to zero, and nobody could say why that was legal. George Berkeley's question in The Analyst (1734) still names the problem better than any textbook: "May we not call them ghosts of departed quantities?"

The story runs in three steps:

• The objection. Newton's method for a rate of change divided by a small increment and then let the increment vanish. George Berkeley (1685 to 1753, Ireland) pointed out that the increment was treated as nonzero while dividing and as zero afterwards. MacTutor notes that his criticisms were well founded and pushed mathematicians toward a logical clarification of calculus.

• The repair. Augustin-Louis Cauchy (1789 to 1857, France) wrote the Cours d'analyse in 1821 for students at the École Polytechnique, building the basic theorems of calculus on limits as rigorously as he could.

• The final form. Karl Weierstrass (1815 to 1897, Germany) and his school later turned "as close as you like" into the epsilon and delta inequalities used today, which is precisely what makes the cancellation rule on this page a theorem rather than a hope.

That is where this page leads. The limit definition of derivative is one big $\frac{0}{0}$ limit, and every first-principles derivative is evaluated with exactly the factoring, conjugate and common-denominator moves above (Example 4 already computed one).

Where Are Algebraic Limit Techniques Used In Real Life?

Algebraic limits are used wherever a quantity is defined at an instant or at a single point where the formula breaks:

• Instantaneous speed. A dropped stone falls $s(t) = 4.9t^2$ metres in $t$ seconds. Its speed at $t = 2$ is $\lim_{h \to 0} \frac{4.9(2 + h)^2 - 4.9(4)}{h} = \lim_{h \to 0} (19.6 + 4.9h) = 19.6,\text{m/s}$ after factoring out $h$. This is the derivative as rate of change, computed with nothing more than this page's algebra.

• Signal processing. The function $\frac{\sin x}{x}$ (called sinc) describes how filters and antennas respond. It is undefined at $x = 0$, and engineers fill that point with the limit value $1$.

• Economics. Marginal cost is the limit of the extra cost per extra unit as the change in units shrinks, the same $\frac{0}{0}$ structure.

• Asymptotic behaviour. Dividing by the highest power tells an engineer the long-run level of a response such as $\frac{2t^2 - 3t + 1}{5t^2 + 4}$, without plotting it.

The Mistakes Students Make Most Often

Most errors in evaluating limits algebraically come from treating $\frac{0}{0}$ as an answer, or from dropping the condition $x \neq a$ that makes the rewrite legal.

Mistake 1: Treating 0/0 As An Answer

Where it slips in: Right after the first substitution, when the result looks finished.

Don't do this: Write $\frac{0}{0} = 1$, $\frac{0}{0} = 0$, or "the limit does not exist" and stop.

The correct way: Read $\frac{0}{0}$ as a signal to rewrite. The rusher writes "DNE" here and moves on, which is why Example 2 shows the table: the true value was $12$.

Mistake 2: Cancelling Terms Instead Of Factors

Where it slips in: Rational expressions such as $\frac{x^2 + x - 6}{x - 2}$, where $x$ appears on both lines.

Don't do this: Cross out the $x$ in $x - 2$ against an $x$ in the numerator. Only whole factors cancel, never pieces of a sum.

The correct way: Factor completely first, then cancel a common factor such as $(x - 2)$. The first instinct with a fraction like this is to "simplify" by striking matching letters, and factoring before cancelling is the habit that removes that error.

Mistake 3: Losing The Sign With Square Roots At Negative Infinity

Where it slips in: Limits as $x \to -\infty$ that contain $\sqrt{x^2 + \ldots}$.

Don't do this: Replace $\sqrt{x^2}$ with $x$. That gives $3$ instead of $-3$ in Example 5.

The correct way: Use $\sqrt{x^2} = |x|$, which equals $-x$ for negative $x$. The memorizer who learned "divide by the highest power" as a rule, without the reason, applies it perfectly and still gets the sign wrong here.

Mistake 4: Splitting A Limit Into Pieces That Do Not Exist

Where it slips in: Complex fractions and differences, such as $\frac{1}{x(x+2)} - \frac{1}{2x}$.

Don't do this: Apply the limit law for a difference when the separate limits are infinite.

The correct way: Combine first. The limit laws only apply when each separate limit exists as a real number.

Wait, one refinement on Mistake 1: "does not exist" is sometimes the correct final answer, as the absolute value example showed. The error is reaching it from $\frac{0}{0}$ alone, before any rewrite.

The software version of Mistake 1 has happened at sea. On 21 September 1997, a crew member on the USS Yorktown entered a zero into a database field, the ship's software attempted a division by zero, and the failure brought down the network and left the cruiser without propulsion. A program that met $\frac{0}{0}$ and stopped had no way to recover; a mathematician who meets it rewrites the expression.

Practice Problems On Evaluating Limits Algebraically

1. Evaluate $\lim_{x \to 4} \dfrac{x^2 - 16}{x - 4}$.

2. Evaluate $\lim_{x \to 1} \dfrac{x^3 - 1}{x^2 - 1}$.

3. Evaluate $\lim_{x \to 0} \dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$.

4. Evaluate $\lim_{x \to \infty} \dfrac{4x^3 - x}{2x^3 + 7x^2}$.

5. Evaluate $\lim_{x \to 0} \dfrac{\sin 4x}{\tan 2x}$.

6. Evaluate $\lim_{x \to -1} \dfrac{\frac{1}{x + 2} - 1}{x + 1}$.

Answer to Question 1: $8$ (difference of squares, then $x + 4$ at $x = 4$).

Answer to Question 2: $\frac{3}{2}$ (factor both: $\frac{x^2 + x + 1}{x + 1}$ at $x = 1$).

Answer to Question 3: $1$ (conjugate gives $\frac{2}{\sqrt{1 + x} + \sqrt{1 - x}}$).

Answer to Question 4: $2$ (same degree, ratio of leading coefficients).

Answer to Question 5: $2$ (write as $\frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \cos 2x \cdot 2$).

Answer to Question 6: $-1$ (common denominator gives $\frac{-1}{x + 2}$).

What To Remember About Evaluating Limits Algebraically

• Evaluating limits algebraically always starts with direct substitution, and a real-number result is the answer.

• A $\frac{0}{0}$ result means rewrite, while $\frac{k}{0}$ means check the sign on each side.

• Factoring, conjugates, common denominators, dividing by the highest power and the $\frac{\sin u}{u}$ limit cover most first-course problems.

• Cancelling is legal because a limit ignores the single input $x = a$.

• Keep "lim" on every line, cancel factors only, and remember $\sqrt{x^2} = |x|$.

Where To Go From Here

Work through Practice Problems 1 to 6 in order. If Problem 3 trips you up, return to "How Do You Rationalize To Evaluate A Limit?"; if Problem 4 does, reread the section on dividing by the highest power and check Example 5 for the sign rule.

Want a teacher to walk through the next set with you? Book a free demo class with a Bhanzu trainer.

Read More

• Limit formulas for algebraic, exponential and trigonometric expressions

• How limits lead to derivatives

• Types of discontinuity: removable, jump and infinite

• Continuity and differentiability together