B.5 Use Algebra to Find a Limit

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.

Show/Hide Step 1 Solution

\[ \lim_{x \to 0} \,\dfrac{(x-2)^2 – 4}{x} = \dfrac{(2-2)^2}{0}=\dfrac{0}{0} \] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. We thus have more work to do. So on to Step 2…

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Step 2. Use algebra and simplify.
In this case, expand the quadratic and then simplify:

Show/Hide Step 2 Solution

\begin{align*}
\lim_{x \to 0}\dfrac{(x-2)^2 – 4}{x} &= \lim_{x \to 0}\dfrac{(x^2 -4x + 4) – 4}{x} \\[8px] &= \lim_{x \to 0}\dfrac{x^2 – 4x}{x} \\[8px] &= \lim_{x \to 0}\dfrac{x(x – 4)}{x} \\[8px] &= \lim_{x \to 0}\dfrac{\cancel{x}(x-4)}{\cancel{x}} \\[8px] &= \lim_{x \to 0}(x – 4) \\[8px] \end{align*}

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Step 3. Substitution to finish.

Show/Hide Step 3 Solution

\begin{align*}
\hphantom{\lim_{x \to 0}\dfrac{(x-2)^2 – 4}{x} } & \hphantom{ = \lim_{x \to 0}\dfrac{(x^2 -4x + 4) – 4}{x}} \\
&= 0 – 4 \\[8px] &= -4 \quad \cmark
\end{align*}

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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, except that the original function is not defined at $x=0,$ whereas the rewritten function is.
How the tactic of using algebra to find a limit works: The left-hand graph has a hole in it at x=0, since the function is undefined there. The right-hand graph does not, since it is defined everywhere.

How the tactic of using algebra to find a limit works: The left-hand graph has a hole in it at x=0, since the function is undefined there. The right-hand graph does not, since it is defined everywhere.

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.

Show/Hide Step 1 Solution

\[ \lim_{x \to 4} \, \frac{x^{-1} – 4^{-1}}{x-4} =\dfrac{4^{-1} – 4^{-1}}{0} =\dfrac{0}{0} \] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is.

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Step 2. Use algebra and simplify.
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.

Show/Hide Step 2 Solution

\begin{align*}
\lim_{x \to 4} \frac{x^{-1} – 4^{-1}}{x-4} &= \lim_{x \to 4} \frac{\frac{1}{x} – \frac{1}{4}}{x-4} \\[8px] &= \lim_{x \to 4} \frac{\frac{4}{4x} – \frac{x}{4x}}{x-4} \\[8px] &= \lim_{x \to 4} \frac{\frac{4-x}{4x}}{x-4} \\[8px] &= \lim_{x \to 4} \left( \frac{4-x}{4x} \right) \left( \frac{1}{x-4} \right) \\[8px] \end{align*}
We can cancel some terms if we factor a $-1$ out of the $4-x:$
\begin{align*}
\hphantom{\lim_{x \to 4} \frac{x^{-1} – 4^{-1}}{x-4} } & \hphantom{ \lim_{x \to 4} \frac{\frac{1}{x} – \frac{1}{4}}{x-4}} \\
&= \lim_{x \to 4} \left( \frac{-(x-4)}{4x} \right) \left( \frac{1}{x-4} \right) \\[8px] &= \lim_{x \to 4} \left( \frac{-\cancel{(x-4})}{4x} \right) \left( \frac{1}{\cancel{(x-4)}} \right) \\[8px] &= \lim_{x \to 4} \frac{-1}{4x}
\end{align*}

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Step 3. Substitution to finish.

Show/Hide Step 3 Solution

\begin{align*}
\hphantom{\lim_{x \to 4} \frac{x^{-1} – 4^{-1}}{x-4} } & \hphantom{ \lim_{x \to 4} \frac{\frac{1}{x} – \frac{1}{4}}{x-4}} \\
&= \frac{-1}{4(4)} \\[8px] &= \frac{-1}{16} \quad \cmark
\end{align*}

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

Practice Problems: Use Algebra to Find a Limit

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}

Show/Hide Solution

We first try Substitution: \[\lim_{x \to 0} \frac{x^2 – 5x}{x} \overset{?}{=} \frac{0 – 0}{0} = \frac{0}{0} \] Once again we have an inderminate result, and so have more work to do.

It may not be immediately obvious what to do, but we do notice that we can factor an x out of the terms in the numerator. So let’s start with that: \begin{align*} \lim_{x \to 0} \frac{x^2 – 5x}{x} &= \lim_{x \to 0}\frac{x(x-5)}{x} \\[8px] &= \lim_{x \to 0}\frac{\cancel{x}(x-5)}{\cancel{x}} \quad \text{[Ah: we can cancel the $x$’s]} \\[8px] &= \lim_{x \to 0}(x-5) \\[8px] &= -5 \implies \quad\text{ (B)} \quad \cmark \end{align*}

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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}

Show/Hide Solution

As always, we first try Substitution: $$\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} \overset{?}{=} \dfrac{(0-5)^2 -25}{0} = \dfrac{25-25}{0} = \dfrac{0}{0}$$ Since the limit is in the form $\dfrac{0}{0},$ it is indeterminate—we don’t yet know what is it. So we have to do some work to turn the expression into a different form that’s more helpful. Let’s try the simple move of expanding the quadratic in the numerator, and then see what happens: \begin{align*} \lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} &= \lim_{h \to 0}\dfrac{(h^2 -10h + 25) – 25}{h} \\[8px] &= \lim_{h \to 0}\dfrac{h^2 – 10h}{h} \\[8px] &= \lim_{h \to 0}\dfrac{h(h – 10)}{h} \\[8px] &= \lim_{h \to 0}\dfrac{\cancel{h}(h-10)}{\cancel{h}} \quad \text{[Cool: cancellation.]}\\[8px] &= \lim_{h \to 0}(h – 10) \\[12px] &= 0\, – 10 \\[8px] &= -10 \implies \quad\text{ (B)} \quad \cmark \end{align*} Notice that the problematic h in the denominator cancelled out, just as we were hoping would happen. We can then use simple Substitution to finish.

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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}

Show/Hide Solution

We first try Substitution: $$\lim_{h \to 0}\frac{(5+h)^{-1} -5^{-1}}{h} = \frac{(5+0)^{-1} – 5^{-1}}{0} = \frac{0}{0} $$ Since this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. We have more work to do. So let’s put the terms in the numerator over a common denominator, and go from there: \begin{align*} \lim_{h \to 0}\frac{(5+h)^{-1} -5^{-1}}{h} &= \lim_{h \to 0}\frac{\dfrac{1}{5+h} -\dfrac{1}{5}}{h} \\[8px] &= \lim_{h \to 0}\frac{\dfrac{5}{5(5+h)} -\dfrac{(5+h)}{5(5+h)}}{h} \\[8px] &= \lim_{h \to 0}\frac{\dfrac{5 – (5+h)}{5(5+h)}}{h} \\[8px] &= \lim_{h \to 0}\frac{\dfrac{-h}{5(5+h)}}{h} \\[8px] &= \lim_{h \to 0}\frac{-h}{5(5+h)h} \\[8px] &= \lim_{h \to 0}\frac{-\cancel{h}}{5(5+h)\cancel{h}} \\[8px] &= \lim_{h \to 0}\frac{-1}{5(5+h)} \\[8px] &= \frac{-1}{5(5+0)} \\[8px] &= \frac{-1}{25} \implies \quad\text{ (D)} \quad \cmark \end{align*}

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Practice Problem #4

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

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.

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

Show/Hide Solution

Let’s first try Substitution: \[ \lim_{h \to 0}\frac{(h-1)^3 + 1}{h} \overset{?}{=} \frac{(-1)^3 +1}{0} = \frac{0}{0} \] Once again, the indeterminate result $\dfrac{0}{0}.$ We have more work to do: Recall (or if needed, see the hint above) \[(h-1)^3 = h^3 -3h^2 + 3h -1\] Then \begin{align*} \lim_{h \to 0}\dfrac{(h-1)^3 + 1}{h} &= \lim_{h \to 0}\dfrac{(h^3 -3h^2 + 3h -1) + 1}{h} \\[8px] &= \lim_{h \to 0}\dfrac{h^3 -3h^2 + 3h}{h} \\[8px] &= \lim_{h \to 0}\dfrac{h(h^2 -3h + 3)}{h} \\ \\ &= \lim_{h \to 0}\dfrac{\cancel{h}(h^2 -3h + 3)}{\cancel{h}} \\[8px] &= \lim_{h \to 0} \left( h^2 – 3h + 3\right) \\[8px] &= 0 – 0 + 3 \\[8px] &= 3 \implies \quad\text{ (C)} \quad \cmark \end{align*}

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The Upshot

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.

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