Calculus · Functions and limits · Lesson 2

Limit algebra

Most limits you’ll meet fall to one of four moves: plug in, factor, combine fractions, or multiply by the conjugate. This lesson drills all four, with extra time on the fraction move from last lesson’s stretch question.

Move 0: just plug in

Limits behave politely with arithmetic. The limit of a sum is the sum of the limits, and the same goes for differences, products, and quotients (as long as the bottom isn’t heading to 0). For polynomials and most everyday functions, that means you can simply substitute.

$$\lim_{x\to 2}(x^3 - 4x + 1) = 8 - 8 + 1 = 1$$

$\displaystyle\lim_{x\to 3}(x^2 + 2x - 1)$

Plug in: $9 + 6 - 1 = 14$.

$\displaystyle\lim_{x\to 2}\frac{x+1}{x-3}$

The bottom heads to $-1$, not 0, so plug in: $\frac{3}{-1} = -3$.

Reading the result of plugging in

a numberThat’s the limit. Done. nonzero / 0The function blows up near the point. No finite limit (lesson 4 handles these). 0 / 0Indeterminate. Something on top and bottom is cancelling to zero together. Find it with algebra, cancel it, plug in again.

The three algebra moves below are three ways of making the hidden common factor $(x - a)$ show itself.

Move 1: factor

If top and bottom are polynomials that both vanish at $x = a$, both have a factor of $(x - a)$. Tap through:

$\displaystyle\lim_{x\to -2}\frac{x^2+5x+6}{x+2}$

  1. Plug in: $\frac{4 - 10 + 6}{0} = \frac00$. So simplify.

  2. Factor the top: two numbers that multiply to 6 and add to 5 are 2 and 3, so $x^2+5x+6 = (x+2)(x+3)$.

  3. Cancel: $\frac{(x+2)(x+3)}{x+2} = x+3$, valid because $x \ne -2$ inside a limit.

  4. Plug in again: $-2 + 3 = 1$.

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

$x^2 - x - 6 = (x-3)(x+2)$ (numbers multiplying to −6 and adding to −1: −3 and 2). Cancel to get $x + 2 \to 5$.

Move 2: combine fractions

This is the move from last lesson’s stretch question. When the top is a difference of fractions, the factor you need is hidden inside them. Merge them into one fraction and it appears.

$\displaystyle\lim_{x\to 3}\frac{\frac1x - \frac13}{x-3}$

  1. Plug in: $\frac{1/3 - 1/3}{0} = \frac00$. Simplify.

  2. Common denominator $3x$: $\frac1x = \frac{3}{3x}$ and $\frac13 = \frac{x}{3x}$, so the top is $\frac{3-x}{3x}$.

  3. Flip the sign to expose the factor: $3 - x = -(x-3)$, so the top is $-\frac{x-3}{3x}$.

  4. Divide by $(x-3)$, which means multiply by $\frac{1}{x-3}$: $\;-\frac{x-3}{3x}\cdot\frac{1}{x-3} = -\frac{1}{3x}$.

  5. Plug in: $-\frac{1}{9}$.

Step 2 only: combine $\frac1x - \frac14$ into a single fraction. Type it like (a-b)/(cx).

Common denominator $4x$: $\frac{4}{4x} - \frac{x}{4x} = \frac{4-x}{4x}$.

Now the whole thing: $\displaystyle\lim_{x\to 4}\frac{\frac1x - \frac14}{x-4}$. (Decimal is fine.)

$\frac{4-x}{4x} = -\frac{x-4}{4x}$. Dividing by $(x-4)$ leaves $-\frac{1}{4x} \to -\frac{1}{16} = -0.0625$. It matches the pattern $-\frac{1}{a^2}$.

Move 3: the conjugate

Square roots hide the factor differently. The trick is the identity $(a-b)(a+b) = a^2 - b^2$: multiplying a root expression by its conjugate (same terms, opposite sign) squares the root away.

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

  1. Plug in: $\frac{2-2}{0} = \frac00$.

  2. Multiply top and bottom by the conjugate $\sqrt{x+4}+2$. That is multiplying by 1, so nothing changes:

    $$\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}$$

  3. Top: $(\sqrt{x+4})^2 - 2^2 = x + 4 - 4 = x$. The root is gone.

  4. Now it’s $\frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{\sqrt{x+4}+2}$. Cancel the $x$.

  5. Plug in: $\frac{1}{2+2} = \frac14$.

$\displaystyle\lim_{x\to 9}\frac{\sqrt{x}-3}{x-9}$ (decimal is fine)

Multiply by $\frac{\sqrt x + 3}{\sqrt x + 3}$. Top: $x - 9$. So the expression is $\frac{x-9}{(x-9)(\sqrt x+3)} = \frac{1}{\sqrt x + 3} \to \frac16 \approx 0.1667$.

Shortcut: $x - 9 = (\sqrt x - 3)(\sqrt x + 3)$, so you could also factor the bottom directly.

Check: pick the move

$\displaystyle\lim_{x\to 2}\frac{x^2+1}{x-2}$. What happens when you plug in, and what’s the limit?

Top heads to 5, bottom to 0. A nonzero number over something tiny is huge, so the function blows up. No algebra will rescue it, because there is no common factor to cancel.

Which move fits $\displaystyle\lim_{x\to 0}\frac{\frac{1}{x+2}-\frac12}{x}$?

A difference of fractions on top means combine them. Common denominator $2(x+2)$: the top is $\frac{2 - (x+2)}{2(x+2)} = \frac{-x}{2(x+2)}$. Divide by $x$: $-\frac{1}{2(x+2)} \to -\frac14$.

Finish that one: $\displaystyle\lim_{x\to 0}\frac{\frac{1}{x+2}-\frac12}{x}$

$-\frac{1}{2(x+2)}$ at $x = 0$ is $-\frac14 = -0.25$. (That is the slope of $\frac{1}{x+2}$ at $x=0$, again the $-\frac{1}{a^2}$ pattern with $a = 2$.)

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

Difference of squares: $(x-1)(x+1)$. Cancel to get $x - 1 \to -2$.

In one sentence: why are we allowed to cancel $(x-a)$ from top and bottom when it would be dividing by zero at $x = a$?

Because a limit never evaluates at $x = a$ itself. For every $x$ near $a$ (but not equal to it), $(x-a) \ne 0$, so cancelling is ordinary algebra.