Limits by Direct Substitution

**Direct substitution** means finding a limit by simply plugging the value into the function: $$\lim_{x \to a} f(x) = f(a)$$ This works for every **polynomial** and for every **rational function** whose denominator is **not zero** at $x = a$. Their graphs have no gaps there, so the value the function approaches is the value it actually takes. **Why it works: the limit laws.** If $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist, then limits can be split up: - sum and difference: $\lim (f \pm g) = \lim f \pm \lim g$ - constant multiple: $\lim (c \cdot f) = c \cdot \lim f$ - product: $\lim (f \cdot g) = \lim f \cdot \lim g$ - quotient: $\lim \frac{f}{g} = \frac{\lim f}{\lim g}$, only if $\lim g \neq 0$ - power: $\lim f^n = (\lim f)^n$ **What the substitution tells you:** - a number, like $\frac{5}{5} = 1$: that number is the limit - $\frac{\text{nonzero}}{0}$, like $\frac{3}{0}$: no finite limit (the graph has a vertical asymptote, see the lesson Infinite Limits) - $\frac{0}{0}$: the **indeterminate form 0/0**. It is NOT an answer. It tells you to rewrite the expression first. **Fixing 0/0:** 1. **Factor and cancel**: both parts share a factor $(x - a)$. Cancel it, then substitute again. 2. **Rationalize**: with a square root like $\sqrt{x} - 2$, multiply top and bottom by the **conjugate** $\sqrt{x} + 2$, cancel, then substitute again. Cancelling is allowed because the limit only looks at $x$ values **near** $a$, never at $x = a$ itself, so $x - a \neq 0$.

Limit of a Polynomial

Find $\lim_{x \to 3} (x^2 - 2x + 3)$.

Check the type of function: $x^2 - 2x + 3$ is a polynomial, defined for every $x$ = Direct substitution works

Substitute $x = 3$: $3^2 - 2 \cdot 3 + 3 = 9 - 6 + 3$ = $6$

Write the limit: $\lim_{x \to 3} (x^2 - 2x + 3) = 6$ = The limit is $6$

Answer: $\lim_{x \to 3} (x^2 - 2x + 3) = 6$. On the first graph the point $(3, 6)$ lies on the curve and the table values close in on $6$.

Rational Function, Denominator Not Zero

Find $\lim_{x \to 2} \frac{x^2 + 1}{x + 3}$.

Check the denominator at $x = 2$: $2 + 3 = 5 \neq 0$ = The quotient law applies

Substitute in the numerator: $2^2 + 1 = 5$ = $5$

Divide: $\frac{5}{5} = 1$ = $1$

Answer: $\lim_{x \to 2} \frac{x^2 + 1}{x + 3} = 1$

0/0: Factor and Cancel

Find $\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3}$.

Try direct substitution: $\frac{9 - 3 - 6}{3 - 3} = \frac{0}{0}$ = Indeterminate form: rewrite first

Factor the numerator: $x^2 - x - 6 = (x - 3)(x + 2)$ = Common factor $(x - 3)$

Cancel the common factor ($x \neq 3$ near the limit): $\frac{(x - 3)(x + 2)}{x - 3} = x + 2$ = $x + 2$

Substitute again: $3 + 2$ = $5$

Answer: $\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3} = 5$. The second graph shows a hole at $(3, 5)$: the function is undefined at $3$, but the limit is $5$.

0/0: Factor Top and Bottom

Find $\lim_{x \to -1} \frac{x^2 + 4x + 3}{x^2 - 1}$.

Try direct substitution: $\frac{1 - 4 + 3}{1 - 1} = \frac{0}{0}$ = Indeterminate form

Factor both parts: $\frac{(x + 1)(x + 3)}{(x + 1)(x - 1)}$ = Common factor $(x + 1)$

Cancel: $\frac{x + 3}{x - 1}$ = Simplified expression

Substitute $x = -1$: $\frac{-1 + 3}{-1 - 1} = \frac{2}{-2}$ = $-1$

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

0/0 with a Square Root: Rationalize

Find $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$.

Try direct substitution: $\frac{\sqrt{4} - 2}{4 - 4} = \frac{0}{0}$ = Indeterminate form

Multiply top and bottom by the conjugate $\sqrt{x} + 2$: $\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)}$ = Top becomes $x - 4$

Cancel $(x - 4)$: $\frac{1}{\sqrt{x} + 2}$ = Simplified expression

Substitute $x = 4$: $\frac{1}{\sqrt{4} + 2} = \frac{1}{4}$ = $\frac{1}{4} = 0.25$

Answer: $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{4}$. The third graph has a hole at $(4, 0.25)$.

In the lessons Evaluating Limits Graphically and Evaluating Limits from Tables you estimated limits by looking. Direct substitution gives the **exact** value in seconds, and it is the method you will use most often in calculus. - **Derivatives**: every derivative starts as a limit of the form $\frac{0}{0}$, for example $\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}$. Factoring and cancelling is exactly how you solve it. - **Physics**: instantaneous speed is the limit of average speeds, found by factoring a difference quotient. - **Continuity**: a function is continuous at $a$ exactly when direct substitution works there (lesson Continuity). - **Holes in graphs**: the factor you cancel shows where a rational function has a hole, and the limit tells you its height. Knowing when you may substitute, and what to do when you get $\frac{0}{0}$, saves you from wrong answers like "the limit is 0" or "the limit does not exist".

Mistake: Stopping at $\frac{0}{0}$ and answering "the limit is 0", "1" or "does not exist"

Why: $\frac{0}{0}$ is not a number and not a final answer. It only says that the top and bottom share a factor that hides the real value.

Correct: Treat $\frac{0}{0}$ as a signal: factor and cancel (or rationalize), then substitute again. $\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3} = 5$.

Mistake: Cancelling terms instead of factors, like $\frac{x^2 - 9}{x - 3} = \frac{x - 9}{-3}$

Why: You may only cancel a factor that multiplies the whole numerator and the whole denominator. $x$ inside a sum is not such a factor.

Correct: Factor first: $\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3$, so the limit as $x \to 3$ is $6$.

Mistake: Treating $\frac{3}{0}$ like $\frac{0}{0}$ and trying to factor

Why: When only the denominator is 0, there is no common factor to cancel. The values grow without bound near that $x$.

Correct: $\frac{\text{nonzero}}{0}$ means there is no finite limit (vertical asymptote). Only $\frac{0}{0}$ calls for factoring or rationalizing.

Mistake: Multiplying only the numerator by the conjugate

Why: Multiplying just one part changes the value of the fraction. You are allowed to multiply by $\frac{\sqrt{x} + 2}{\sqrt{x} + 2}$, which equals 1.

Correct: Multiply top AND bottom by the conjugate, and leave the bottom factored so you can cancel $(x - 4)$.

Instantaneous Speed

A speedometer shows speed at one instant. Physicists find it as the limit of average speeds over shorter and shorter time intervals, and that limit always starts as 0/0.

A cart moves so that its position is $s(t) = t^2 + 3t$ meters after $t$ seconds. Its average speed between second 2 and time $t$ is $\frac{s(t) - s(2)}{t - 2} = \frac{t^2 + 3t - 10}{t - 2}$. Factoring gives $\frac{(t - 2)(t + 5)}{t - 2} = t + 5$.

Growing Square Tiles

Designers and engineers ask how fast an area changes when a length changes. For a square of side $x$ the area is $x^2$, and the rate of change at one size is a 0/0 limit.

A square tile with side $x$ cm has area $A(x) = x^2$ cm². When the side grows from 5 cm to $x$ cm, the area changes on average by $\frac{x^2 - 25}{x - 5}$ cm² per cm of side.

Plant Growth Rate

Biologists model growth with square root formulas. The growth rate on a given day is a limit that needs rationalizing.

A seedling is $h(t) = \sqrt{t}$ dm tall after $t$ days. Its average growth between day 9 and day $t$ is $\frac{\sqrt{t} - 3}{t - 9}$ dm per day.

For polynomials, $\lim_{x \to a} f(x) = f(a)$: just substitute

For rational functions, substitute when the denominator is not 0 at $x = a$

The limit laws let you split a limit into sums, products, quotients (denominator limit not 0) and powers

$\frac{0}{0}$ is an indeterminate form, never the answer: rewrite the expression first

Factor and cancel the common factor $(x - a)$, or multiply by the conjugate when there is a square root, then substitute again

$\frac{\text{nonzero}}{0}$ means no finite limit (vertical asymptote), not a hole

Q: Why may I cancel $(x - 3)$ if it is 0 at $x = 3$?

A: A limit as $x \to 3$ only uses $x$ values close to 3, never $x = 3$ itself. For all those values $x - 3 \neq 0$, so cancelling is a legal step. The two expressions agree everywhere except at $x = 3$, and the limit does not care about that single point.

Q: If I get $\frac{0}{0}$, does the limit exist?

A: You cannot tell yet. $\frac{0}{0}$ can hide any value: $\lim_{x \to 0} \frac{2x}{x} = 2$, $\lim_{x \to 0} \frac{x^2}{x} = 0$. Simplify first, then decide.

Q: When should I rationalize instead of factoring?

A: When the 0 comes from a difference with a square root, like $\sqrt{x} - 2$ or $\sqrt{x + 4} - 2$. Multiply top and bottom by the conjugate (same terms, opposite sign in the middle), because $(\sqrt{x} - 2)(\sqrt{x} + 2) = x - 4$ has no root left.

Q: Is the limit the same as the function value?

A: For polynomials and for rational functions where the denominator is not 0, yes: that is exactly what direct substitution says. At a hole, the function value does not exist, but the limit does (lessons Continuity and Types of Discontinuity go deeper).

Exit Presentation

Limits by Direct Substitution

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Pre-Calculus / Limits

Limits by Direct Substitution

Find limits algebraically: plug in the value when you can, and when you get 0/0, factor and cancel or rationalize first.

Advanced30 min

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