Evaluating Limits by Direct Substitution
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.3: “The Limit Laws” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
| Supplementary | OpenStax Calculus Volume 1, Section 2.4: “Continuity” |
Both sources are free and openly licensed.
Key idea
The fastest way to evaluate a limit is often the most obvious way: plug in the number and compute.
If $f$ is a “nice” function at $x = a$ -- meaning it is defined there, its graph has no break, and nearby values do not do anything unexpected -- then the limit equals the function value. This is not a coincidence; it is the definition of continuity at a point. Direct substitution is really asking: “Is $f$ continuous at $a$?” If yes, the limit is $f(a)$. If not, you need a different strategy.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Direct Substitution Property. If $f$ is a polynomial, a rational function with nonzero denominator at $a$, or a composition of such functions that remains defined at $a$, then \[ \lim_{x \to a} f(x) = f(a). \]
When to use it. Always try direct substitution first. If the result is a specific finite number, you are done. If the result is $\frac{0}{0}$, $\frac{\infty}{\infty}$, or another indeterminate form, substitution has told you something (the technique fails here), and you must use algebra to proceed.
Key Concepts
1. Why Direct Substitution Works for Polynomials
A polynomial is a sum of terms of the form $c \cdot x^n$. For any real number $a$, the limit of $x^n$ as $x \to a$ is $a^n$ -- this follows from the basic limit law $\lim_{x \to a} x = a$ applied repeatedly using the product rule for limits. Adding finitely many such terms and applying the sum rule gives $\lim_{x \to a} p(x) = p(a)$ for every polynomial $p$.
In other words, polynomials are continuous everywhere. You can always plug in.
| Function | $a$ | $\lim_{x \to a} f(x)$ | Method |
|---|---|---|---|
| $f(x) = 3x^2 - 5x + 2$ | $4$ | $3(16) - 5(4) + 2 = 30$ | plug in |
| $f(x) = x^3 - 1$ | $2$ | $8 - 1 = 7$ | plug in |
| $f(x) = -x^4 + 2x$ | $-1$ | $-1 - 2 = -3$ | plug in |
2. Rational Functions: Check the Denominator
A rational function $r(x) = p(x)/q(x)$ is continuous wherever its denominator is nonzero. So:
- If $q(a) \neq 0$: plug in, compute $p(a)/q(a)$.
- If $q(a) = 0$ and $p(a) \neq 0$: the limit does not exist (infinite or one-sided infinite).
- If $q(a) = 0$ and $p(a) = 0$: you get $\frac{0}{0}$, an indeterminate form; substitution fails, and you must factor or rationalize.
Example 1. Evaluate $\displaystyle\lim_{x \to 3} \frac{x^2 + 1}{x - 1}$.
Check: denominator at $x = 3$ is $3 - 1 = 2 \neq 0$. Safe to plug in: \[ \frac{9 + 1}{3 - 1} = \frac{10}{2} = 5. \]
Example 2. Evaluate $\displaystyle\lim_{x \to -2} \frac{x^3 - 4x}{x^2 + 3x + 2}$.
Factor the denominator: $x^2 + 3x + 2 = (x+1)(x+2)$. At $x = -2$: denominator is $(−1)(0) = 0$. Check numerator: $(-2)^3 - 4(-2) = -8 + 8 = 0$. Indeterminate form $\frac{0}{0}$.
Direct substitution fails. (This is the signal that you need algebraic techniques; see the lesson on Algebraic Techniques for Limits.)
3. Trigonometric and Root Functions
The functions $\sin$, $\cos$, $\tan$, $\sqrt{\cdot}$, and their combinations are continuous on their natural domains. Direct substitution applies wherever the function is defined.
Example 3. Evaluate $\displaystyle\lim_{x \to \pi/4} \sin(x) \cdot \cos(x)$.
Both sine and cosine are continuous everywhere. Plug in: \[ \sin\!\left(\frac{\pi}{4}\right) \cdot \cos\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{4} = \frac{1}{2}. \]
Example 4. Evaluate $\displaystyle\lim_{x \to 4} \sqrt{x - 3}$.
The square root function is continuous wherever its argument is positive. At $x = 4$: $x - 3 = 1 > 0$. Plug in: \[ \sqrt{4 - 3} = \sqrt{1} = 1. \]
Example 5. Evaluate $\displaystyle\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$.
Numerator at $x = 9$: $\sqrt{9} - 3 = 0$. Denominator: $9 - 9 = 0$. Indeterminate form. Direct substitution fails. (Rationalize to resolve; see Algebraic Techniques.)
4. The Connection to Continuity
The reason direct substitution works is exactly continuity. The formal definition says $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$. So “direct substitution works” and “the function is continuous at $a$” are the same statement.
a limit always equals the function value at the point.
This is the limit-equals-function-value error. Direct substitution works precisely when the function IS continuous at $a$, and not in general. Consider the function $f(x) = \dfrac{x^2 - 1}{x - 1}$. At $x = 1$, $f$ is undefined (denominator is zero). Yet the limit exists: $\lim_{x \to 1} f(x) = \lim_{x \to 1}(x + 1) = 2$. The limit is 2 even though $f(1)$ does not exist. A limit is about what nearby values approach; it is not required to equal the value at the point, and the value at the point is not even required to exist. Direct substitution is a useful shortcut when the function is continuous, not a universal rule.
if a function is discontinuous at $a$, its limit at $a$ does not exist.
This is the continuity-vs-limit-existence error, and it is the reverse of the one above. Discontinuity and limit non-existence are different things. A function can be discontinuous at $a$ for three reasons: it is undefined there, the limit does not exist, or the limit exists but does not equal the function value. Only the second reason means the limit fails to exist. For $f(x) = \dfrac{x^2 - 1}{x - 1}$, the function is discontinuous at $x = 1$ (undefined there), but the limit $\lim_{x \to 1} f(x) = 2$ exists. For $g(x) = \dfrac{|x|}{x}$, the function is discontinuous at $x = 0$ and the limit also fails to exist, because the one-sided limits disagree. The two cases look similar but are different: always compute the limit independently of whether the function value exists.
This means:
- Polynomials: continuous everywhere -- always plug in.
- Rational functions: continuous wherever the denominator is nonzero.
- $\sin$, $\cos$: continuous everywhere.
- $\tan$: continuous wherever $\cos \neq 0$, i.e., $x \neq \pi/2 + n\pi$.
- $\sqrt{x}$: continuous for $x > 0$; at $x = 0$ only the right-hand limit exists.
Checking continuity first saves you from attempting substitution at a discontinuity.
5. The Substitution Protocol
A reliable four-step sequence for any limit:
- Try direct substitution. Compute $f(a)$.
- Specific finite number. Done.
- Nonzero over zero (e.g., $\frac{5}{0}$). Limit is $\pm\infty$ or does not exist. Investigate one-sided limits.
- Zero over zero or other indeterminate form. Substitution has failed but told you something. Move to algebraic techniques (factoring, rationalizing, or the squeeze theorem).
Common Errors
| Error | Example | Correction |
|---|---|---|
| Skipping the denominator check | Computing $\frac{f(a)}{0}$ and stopping | Check $q(a)$ before substituting into a rational function |
| Treating $\frac{0}{0}$ as “zero” | Writing $\lim = 0$ because “numerator is zero” | $\frac{0}{0}$ is indeterminate; you must simplify |
| Treating $\frac{0}{0}$ as “undefined” | Writing “DNE” immediately | $\frac{0}{0}$ is indeterminate; the limit may still exist after simplification |
| Forgetting domain restrictions | Substituting $x = 0$ into $\sqrt{x}$ without noting the limit is one-sided | Check whether $a$ is in the domain first |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Evaluate $\displaystyle\lim_{x \to 5} (2x^2 - 3x + 1)$.
Show answer
$2(25) - 3(5) + 1 = 50 - 15 + 1 = 36$.
Problem 2. Evaluate $\displaystyle\lim_{x \to 2} \frac{x^2 + x - 1}{x + 3}$.
Show answer
Denominator at $x = 2$: $2 + 3 = 5 \neq 0$. Plug in: $\dfrac{4 + 2 - 1}{5} = \dfrac{5}{5} = 1$.
Problem 3. Evaluate $\displaystyle\lim_{x \to \pi} \cos(x) + 2$.
Show answer
$\cos(\pi) + 2 = -1 + 2 = 1$.
Level 2 -- Classify the Outcome
Problem 4. For each limit below, state whether direct substitution gives a specific number, $\frac{0}{0}$, or $\frac{c}{0}$ with $c \neq 0$.
(a) $\displaystyle\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$
(b) $\displaystyle\lim_{x \to 3} \frac{x + 2}{x - 3}$
(c) $\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$
Show answers
(a) Numerator $= 1 - 1 = 0$, denominator $= 0$. Indeterminate $\frac{0}{0}$.
(b) Numerator $= 5 \neq 0$, denominator $= 0$. The form $\frac{5}{0}$; the limit does not exist (or is $\pm\infty$).
(c) Numerator $= \sin 0 = 0$, denominator $= 0$. Indeterminate $\frac{0}{0}$. (This limit is 1, but requires the squeeze theorem.)
Problem 5. Evaluate $\displaystyle\lim_{x \to -3} \frac{x^2 - 9}{x + 3}$ using only direct substitution if possible; if not, explain why and state the form you get.
Show answer
Numerator: $9 - 9 = 0$. Denominator: $0$. Direct substitution gives $\frac{0}{0}$. Cannot evaluate by substitution alone. (Factor numerator as $(x-3)(x+3)$, cancel, and get $\lim = -3 - 3 = -6$.)
Level 3 -- Combining Ideas
Problem 6. Evaluate $\displaystyle\lim_{x \to \pi/6} \left(\sin^2 x + \cos x\right)$.
Show answer
Both $\sin$ and $\cos$ are continuous everywhere. Plug in $x = \pi/6$: \[ \sin^2\!\left(\frac{\pi}{6}\right) + \cos\!\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 + \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{\sqrt{3}}{2} = \frac{1 + 2\sqrt{3}}{4}. \]
Problem 7. Find all values of $a$ for which $\displaystyle\lim_{x \to a} \frac{x^2 + 5x + 6}{x^2 - 4}$ can be evaluated by direct substitution.
Show answer
The denominator is $x^2 - 4 = (x-2)(x+2)$. It equals zero when $x = 2$ or $x = -2$. At all other values of $a$, direct substitution applies.
Note: $\lim_{x \to -3}$ works (denominator is $9 - 4 = 5$), but $\lim_{x \to 2}$ and $\lim_{x \to -2}$ require further analysis (numerator $= 0$ there too, so $\frac{0}{0}$ in both cases).
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Think of direct substitution as a diagnostic tool, not just a computation.
When you plug $a$ into $f(x)$ and get a specific number, you have confirmed two things at once: the limit exists and equals $f(a)$, and $f$ is continuous at $a$. A single substitution does double duty.
When you get $\frac{0}{0}$, the limit has not failed -- only the method has failed. The $\frac{0}{0}$ form is a signal: there is likely a shared factor in numerator and denominator that must be canceled before the limit reveals itself.
When you get $\frac{c}{0}$ with $c \neq 0$, the function is blowing up near $a$. Check both sides to determine whether the limit is $+\infty$, $-\infty$, or does not exist because the two sides disagree.
Connections
Within Calculus I (MATH161)
- Limit laws (Stewart 1.6): Direct substitution is a shortcut that combines the limit laws for sums, products, and quotients into a single step when continuity holds.
- Continuity (Stewart 1.8): “Direct substitution works at $a$” is exactly the statement “$f$ is continuous at $a$.” The connection runs both ways.
- Algebraic techniques: When substitution yields $\frac{0}{0}$, the next lesson applies factoring, rationalizing, or multiplying by conjugates to resolve the indeterminate form.
Toward Later Work
- Derivatives: The derivative at $a$ is defined as $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$, which always gives $\frac{0}{0}$ at $h = 0$. Every derivative computation begins exactly where direct substitution fails.
- L’Hopital’s Rule (MATH162): When algebraic manipulation does not resolve $\frac{0}{0}$ or $\frac{\infty}{\infty}$, derivatives provide another path.
Back to Calculus I Skills | Previous: Limit Laws | Next: Algebraic Techniques for Limits