Use product and quotient laws

Width times height

The orange sides of the teal rectangle have lengths \(x + 1\) and \(10 - 2x\), so its area is their product.

Move the slider to bring x toward 2 from either side. The width approaches 3, and the height approaches 6.

At \(x = 1.9\) the sides are 2.9 and 6.2, and the area is 17.98.

As x comes closer to 2, the area comes closer to 18, which is 3 times 6.

Any two sides

As x approaches an input a, the width approaches \(\cy{L}\) and the height approaches \(\cy{M}\), so the dashed rectangle has sides \(\cy{L}\) and \(\cy{M}\).

The green pieces are area that the teal rectangle has beyond the dashed one, and the red pieces are area that it lacks.

Move x toward a. Each piece becomes thinner, because each side comes close to its limit.

The area approaches the area of the dashed rectangle, \(\cy{L}\cdot\cy{M}\), so the limit of a product is the product of the limits.

Notation

1. Let the width be \(f(x)\) and the height be \(g(x)\). Suppose \(f(x)\) approaches \(\cy{L}\) and \(g(x)\) approaches \(\cy{M}\) as \(x\) approaches \(a\). The laws need both limits to exist.

   \[\begin{aligned} \lim_{x \to a} f(x) = \cy{L} \\ \lim_{x \to a} g(x) = \cy{M} \end{aligned}\]

2. The product law says that the limit of a product is the product of the limits.

   \[\lim_{x \to a} f(x)\,g(x) = \cy{L} \cdot \cy{M}\]

3. The quotient law says that the limit of a quotient is the quotient of the limits. The quotient is the number of chalk bars, each of length \(g(x)\), that fit along the orange bar of length \(f(x)\). If \(f(x)\) is close to \(L\) and \(g(x)\) is close to \(M\), this number is close to \(L\) divided by \(M\).

   \[\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\cy{L}}{\cy{M}}\]

4. The quotient law needs \(M\) to be different from 0, so find the limit of the denominator before you divide.

   \[\cy{M} \neq 0\]

When it fails

The denominator \(g(x)\) now approaches 0, so the chalk bars become very short, while \(f(x)\) still approaches a nonzero number. More and more chalk bars fit along the orange bar, so the quotient grows without bound.

The quotient law needs a nonzero limit for the denominator, so it does not apply here, and the quotient has no limit.

Example 1: a product

Find the limit of \((x+2)(3x-5)\) as \(x\) approaches 4.

1. The width \(x+2\) is a sum of \(x\) and a constant, so the sum law gives its limit.

   \[\lim_{x \to 4} (x+2) = \cy{6}\]

2. The height \(3x-5\) is 3 times \(x\) minus 5, so the constant multiple and difference laws give its limit.

   \[\lim_{x \to 4} (3x-5) = \cy{7}\]

3. Both limits exist, so the product law applies. The limit is their product, which is the area of the dashed rectangle.

   \[6 \cdot 7 = \cy{42}\]

Example 2: a zero in the denominator

Find the limit of \(\dfrac{x+4}{x-2}\) as \(x\) approaches 2.

1. Find the limits of the numerator and the denominator. The graph is green where the quotient is positive and red where it is negative.

   \[\begin{aligned} \lim_{x \to 2} (x+4) = 6 \\ \lim_{x \to 2} (x-2) = 0 \end{aligned}\]

2. A tempting wrong move: Divide the limits and call the result infinite.

   \[\dfrac{6}{0} = \infty\]

   The quotient law needs a denominator limit that is not 0, so it does not apply here. To the right of 2 the green curve rises without bound, and to the left the red curve falls without bound. The two sides do not approach one value.

3. Near 2, the numerator is close to 6 and the denominator is close to 0. The quotient is 601 at \(x = 2.01\) and −599 at \(x = 1.99\), so the limit does not exist.

   \[\begin{aligned} \frac{2.01+4}{2.01-2} = 601 \\ \frac{1.99+4}{1.99-2} = -599 \end{aligned}\]

Practice

Find the limit of \(\dfrac{4x-2}{x-3}\) as \(x\) approaches 5.

Find the limit of the denominator first. If it is not 0, the quotient law applies.

Questions

The graph shows \(f\), \(g\) and \(h\) near \(x = 3\). An open circle marks the height that a curve approaches.

1. What is the limit of \(f(x)\,g(x)\) as \(x\) approaches 3?

   • −8
   • 2
   • 8
   Answer

   −8

2. What is the limit of \(\dfrac{f(x)}{h(x)}\) as \(x\) approaches 3?

   • It does not exist
   • 0
   • 4
   • −2
   Answer

   It does not exist

3. What is the limit of \((2x+1)(x-4)\) as \(x\) approaches \(-3\)?

   • 35
   • −12
   • −35
   Answer

   35

Summary

You find the limit of a product by multiplying the limits, and the limit of a quotient by dividing them.

This works because each part stays close to its limit, so the product or the quotient stays close to its limit too.

Use the quotient law only after you confirm that the limit of the denominator is not 0.

This is one of four skills in topic 1.5 of Which & Why: AP Calculus.