As a third and final tactic, let’s look at other ways to use algebra to find a limit. These are just “other algebraic moves” — things like expanding a quadratic, or putting terms over a common denominator. Basically: do what you gotta do to keep modifying the expression until you can use Substitution. We have practice problems below for you to use, of course each with a complete solution.

Rather than providing an Example, let’s dive in with some Scaffolded Problems. Please give them a try, and use the additional guidance in each Step 2 if you’d like.

Use algebra to find a limit: Scaffolded Problem #1

Find $\displaystyle{\lim_{x \to 0} \,\dfrac{(x-2)^2 – 4}{x}}$.

**Solution.**

**Step 1. First try substitution.**

In this case, expand the quadratic and then simplify:

This approach works for essentially the same reason the factoring tactic and the conjugate tactic work: the functions $\dfrac{(x-2)^2-4}{x}$ and $x-4$ are the same,

Use algebra to find a limit: Scaffolded Problem #2

Find $\displaystyle{\lim_{x \to 4} \frac{x^{-1} – 4^{-1}}{x-4}}$.

**Solution.**

**Step 1. First try substitution.**

It’s hard to see what’s going on with the negative exponents, so first write the numerator-terms as fractions. Then put them over a common denominator. Finally, as usual, simplify.

Now use the same types of approaches for the following practice problems.

Practice Problem #1

Find $\displaystyle{\lim_{x \to 0} \dfrac{x^2 - 5x}{x} }.$
\begin{array}{lllll} \text{(A) }0 && \text{(B) }-5 && \text{(C) }-10 && \text{(D) DNE} && \text{(E) None of these} \end{array}

Practice Problem #2

Find $\displaystyle{\lim_{h \to 0}\dfrac{(h-5)^2 - 25}{h}}$.
\begin{array}{lllll} \text{(A) }10 && \text{(B) }-10 && \text{(C) }5 && \text{(D) }-5 && \text{(E) None of these} \end{array}

Practice Problem #3

Find $\displaystyle{\lim_{h \to 0}\frac{(5+h)^{-1} -5^{-1}}{h}}$.
\begin{array}{lllll} \text{(A) }\dfrac{1}{5} && \text{(B) }-\dfrac{1}{5} && \text{(C) }\dfrac{1}{25} && \text{(D) }-\dfrac{1}{25} && \text{(E) None of these} \end{array}

Practice Problem #4

Find $\displaystyle{ \lim_{h \to 0}\dfrac{(h-1)^3 + 1}{h}}$.

\begin{array}{lllll} \text{(A) }1 && \text{(B) }2 && \text{(C) }3 && \text{(D) nonexistent} && \text{(E) None of these} \end{array}

Hint: Don't remember the formula to expand a cubic?

Most people don’t. So go with what you do know: pull out the quadratic, expand it, and then do a few lines of simple algebra.

[Begin initial algebra sub-problem:] \begin{align*} (h-1)^3 &= (h-1)(h-1)^2 \\ \\ &= (h-1)(h^2 -2h + 1) \\ \\ &= h(h^2 -2h + 1) – (h^2 -2h + 1) \\ \\ &= (h^3 -2h^2 + h) – (h^2 – 2h + 1) \\ \\ &= h^3 -3h^2 + 3h -1 \end{align*} With that, you’re ready to proceed with the rest of the solution.

[End sub-problem.]

Especially on an exam, don’t let something like this trip you up.

Just dive in and do the algebra as-needed.

[Begin initial algebra sub-problem:] \begin{align*} (h-1)^3 &= (h-1)(h-1)^2 \\ \\ &= (h-1)(h^2 -2h + 1) \\ \\ &= h(h^2 -2h + 1) – (h^2 -2h + 1) \\ \\ &= (h^3 -2h^2 + h) – (h^2 – 2h + 1) \\ \\ &= h^3 -3h^2 + 3h -1 \end{align*} With that, you’re ready to proceed with the rest of the solution.

[End sub-problem.]

Especially on an exam, don’t let something like this trip you up.

Just dive in and do the algebra as-needed.

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- When you try Substitution, if you obtain $\dfrac{0}{0}$ and have a quadratic (or cubic, or …) you can expand, or have some fractions you can put over a common denominator, do it. After simplifying, Substitution will probably work.

On the next screen we’ll take a quick look at the Squeeze (or “Sandwich”) Theorem, which you should at least know about.

Questions? Comments? Join the discussion over on the Forum!