Matheno - Learn Well and Excel

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.

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.

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

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The Upshot

  1. 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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