Lesson 1.7 · Unit 1 · CED topics 1.7 and 1.9
Selecting procedures and connecting representations
You now own a toolbox: direct substitution, factoring, conjugates, common denominators, trigonometric limits, the squeeze theorem, and estimates from graphs and tables. The College Board treats these two topics (1.7 and 1.9) as skills topics rather than as new content. There is nothing new to memorize here; the goal is to practice choosing the right tool by looking at the structure of a problem, and to move comfortably between the graphical, numerical, and analytical forms of the same limit. On the exam, mixed limit questions are common, and the first mark often goes to the student who picks the right method quickly.
A decision guide
Always try direct substitution first. Then match what you see to a method.
| What you see | What to try |
|---|---|
| A polynomial, root, exponential, logarithm, sine or cosine, with substitution giving a real number | Direct substitution (the function is continuous at \(c\)) |
| Rational function, substitution gives \(\frac{0}{0}\) | Factor and cancel the common \((x-c)\) |
| A difference or sum containing a square root, giving \(\frac{0}{0}\) | Multiply by the conjugate |
| A fraction inside a fraction, giving \(\frac{0}{0}\) | Common denominator, simplify, cancel |
| \(\sin(kx)\) or \(\tan(kx)\) over \(x\) as \(x\to0\), or \(\frac{1-\cos x}{x}\) | Rewrite using \(\frac{\sin u}{u}\to1\) and identities |
| A bounded oscillating factor times something that approaches \(0\) | Squeeze theorem |
| Piecewise function at a boundary | Left and right limits separately |
| Nonzero number over a quantity approaching \(0\) | The function is unbounded (Lesson 1.14) |
| Only a graph or table is given | Read both sides; say "does not exist" if they disagree |
For each limit, choose a method and find the value.
| Limit | Method | Value |
|---|---|---|
| \(\displaystyle\lim_{x\to4}\dfrac{x^2-16}{x-4}\) | factor and cancel | \(\dfrac{(x-4)(x+4)}{x-4}=x+4\to8\) |
| \(\displaystyle\lim_{x\to0}\dfrac{\sqrt{x+1}-1}{x}\) | conjugate | \(\dfrac{1}{\sqrt{x+1}+1}\to\dfrac12\) |
| \(\displaystyle\lim_{x\to0}\dfrac{\sin4x}{x}\) | trig limit | \(4\cdot\dfrac{\sin4x}{4x}\to4\) |
| \(\displaystyle\lim_{x\to0}x^2\cos\dfrac1x\) | squeeze | \(-x^2\le\cdot\le x^2\), so \(0\) |
| \(\displaystyle\lim_{x\to-2}\dfrac{x^2+1}{x+3}\) | substitution | \(\dfrac{5}{1}=5\) |
| \(\displaystyle\lim_{x\to0}\dfrac{\frac{1}{x+1}-1}{x}\) | common denominator | \(\dfrac{-x}{(x+1)x}=\dfrac{-1}{x+1}\to-1\) |
Connecting representations
A limit is one number, but you can see it in three forms. Skills questions ask you to connect them, for example to identify which graph, table, or formula matches a given limit statement, or to re-express a limit given in one form in another.
Consider \(\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}\).
Numerical. At \(x=0.01\) the quotient is \(\dfrac{\sqrt{4.01}-2}{0.01}\approx0.2498\); at \(x=-0.01\) it is \(\dfrac{\sqrt{3.99}-2}{-0.01}\approx0.2502\). Both sides are near \(0.25\).
Graphical. The graph is a smooth curve through height \(0.25\) with a hole at \(x=0\) (the function is undefined there).
Analytical. Multiply by the conjugate: \(\dfrac{(x+4)-4}{x(\sqrt{x+4}+2)}=\dfrac{1}{\sqrt{x+4}+2}\to\dfrac14\).
All three agree: the limit is \(\dfrac14\). The analytic form is the one you would present as a justification; the other two confirm it and help you spot mistakes.
A graph of \(f\) has a hole at \((3,6)\), with the curve approaching height \(6\) from both sides, and a closed circle at \((3,2)\). Write the limit statement and the value statement that describe this.
\(\displaystyle\lim_{x\to3}f(x)=6\) and \(f(3)=2\). The graph shows both facts; they are different statements and neither implies the other.
- Try direct substitution first; if it gives a real number, you are done.
- For \(\frac{0}{0}\), match the structure: factor, conjugate, common denominator, trig limit.
- A bounded factor times something approaching \(0\) suggests the squeeze theorem.
- Move easily between a graph, a table, and a formula for the same limit; use algebra for the justification.
- Name the method in your work and write the equivalent expression before you substitute.
- Evaluate \(\displaystyle\lim_{x\to9}\dfrac{x-9}{\sqrt{x}-3}\) and name the method.
Show answer
Conjugate (or factor \(x-9=(\sqrt x-3)(\sqrt x+3)\)): the expression equals \(\sqrt x+3\) for \(x\ne9\), so the limit is \(6\).
- Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\frac{1}{x+5}-\frac15}{x}\) and name the method.
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Common denominator: the numerator is \(\dfrac{5-(x+5)}{5(x+5)}=\dfrac{-x}{5(x+5)}\), so the quotient is \(\dfrac{-1}{5(x+5)}\to-\dfrac{1}{25}\).
- Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\sin2x}{\sin5x}\).
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Rewrite \(\dfrac{\sin2x}{\sin5x}=\dfrac{\sin2x}{2x}\cdot\dfrac{5x}{\sin5x}\cdot\dfrac25\to1\cdot1\cdot\dfrac25=\dfrac25\).
- Evaluate \(\displaystyle\lim_{x\to-3}\dfrac{x^2+x-6}{x+3}\).
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Factor: \(x^2+x-6=(x+3)(x-2)\), so the expression is \(x-2\) for \(x\ne-3\), and the limit is \(-5\).
- Evaluate \(\displaystyle\lim_{x\to0}x^3\sin\dfrac{1}{x^2}\).
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\(-1\le\sin\dfrac1{x^2}\le1\) and \(|x^3|\) bounds the product: \(-|x|^3\le x^3\sin\dfrac1{x^2}\le|x|^3\). Both bounds approach 0, so the limit is 0 (squeeze theorem).
- The table of \(g(x)=\dfrac{x^2-9}{x-3}\) gives \(5.999\) at \(x=2.999\) and \(6.001\) at \(x=3.001\); the graph shows a hole at \((3,6)\). Which representation would you use to justify \(\displaystyle\lim_{x\to3}g(x)=6\) on a free-response question, and why?
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The algebraic one: \(\dfrac{(x-3)(x+3)}{x-3}=x+3\) for \(x\ne3\), so the limit is \(6\). The table and graph support the result, but a justification needs a mathematical argument.