Finding Limits Algebraically

Tables and graphs estimate; algebra gives the exact answer. The workhorse is direct substitution: for any polynomial, rational, root, trig, exponential, or log function, if plugging in $x=a$ gives a defined number, that number is the limit.

Step 1 is always: just substitute

Try $x=a$ first. One of three things happens:

Clearing the $frac00$ form

$\frac00$ is a signal, not an answer. The hidden factor of $(x-a)$ sits in both top and bottom — cancel it, then substitute again. Three tools do almost all the work:

$\textbf{factor}\quad\bullet\quad\textbf{rationalize (conjugate)}\quad\bullet\quad\textbf{combine fractions}$

Worked example 1

Evaluate $\displaystyle\lim_{x\to3}\,(x^2-5x+2)$.

1. 1.Polynomial ⇒ direct substitution is legal.
2. 2.$3^2-5(3)+2 = 9-15+2$.

Answer: $-4$

Worked example 2

Evaluate $\displaystyle\lim_{x\to2}\frac{x^2-4}{x-2}$.

1. 1.Substitute: $\frac{0}{0}$ — indeterminate.
2. 2.Factor the top: $\dfrac{(x-2)(x+2)}{x-2}$.
3. 3.Cancel $(x-2)$: leaves $x+2$. Now substitute: $2+2$.

Answer: $4$

Worked example 3

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

1. 1.Substitute: $\frac{0}{0}$.
2. 2.Factor: $\dfrac{(x+3)(x-2)}{x+3}=x-2$.
3. 3.Substitute $-3$: $-3-2$.

Answer: $-5$

Worked example 4

Evaluate $\displaystyle\lim_{x\to0}\frac{\sqrt{x+9}-3}{x}$.

1. 1.Substitute: $\frac{0}{0}$. A root ⇒ multiply by the conjugate.
2. 2.$\dfrac{\sqrt{x+9}-3}{x}\cdot\dfrac{\sqrt{x+9}+3}{\sqrt{x+9}+3}=\dfrac{(x+9)-9}{x(\sqrt{x+9}+3)}$.
3. 3.$=\dfrac{x}{x(\sqrt{x+9}+3)}=\dfrac{1}{\sqrt{x+9}+3}$.
4. 4.Substitute $0$: $\dfrac{1}{3+3}$.

Answer: $\dfrac{1}{6}$

Worked example 5

Evaluate $\displaystyle\lim_{x\to0}\frac{\frac{1}{x+4}-\frac14}{x}$.

1. 1.Combine the top: $\dfrac{1}{x+4}-\dfrac14=\dfrac{4-(x+4)}{4(x+4)}=\dfrac{-x}{4(x+4)}$.
2. 2.Divide by $x$ (i.e. multiply by $\frac1x$): $\dfrac{-x}{4(x+4)}\cdot\dfrac1x=\dfrac{-1}{4(x+4)}$.
3. 3.Substitute $0$: $\dfrac{-1}{4(4)}$.

Answer: $-\dfrac{1}{16}$

Worked example 6

Evaluate $\displaystyle\lim_{x\to1}\frac{x^3-1}{x-1}$.

1. 1.$\frac00$. Use the difference of cubes: $x^3-1=(x-1)(x^2+x+1)$.
2. 2.Cancel $(x-1)$: $x^2+x+1$. Substitute $1$: $1+1+1$.

Answer: $3$

Common trap: Cancelling, then forgetting to substitute

After you cancel the common factor you are not finished — you still have to plug $x=a$ into the simplified expression. Students cancel $\frac{x^2-4}{x-2}$ to $x+2$ and stop; the limit is $x+2$ evaluated at $2$, which is $4$.

Hack: The $frac00$ decoder ring

When substitution gives $\frac00$, the fix is dictated by what you see: a polynomial on top or bottom → factor; a square root → multiply by the conjugate; a fraction inside the fraction (a “complex fraction”) → combine the little fractions first. You almost never need to guess.