Combine limits using arithmetic

Two heights, one total

The curve shows \(f(x) = x^2 + 2\), and the line shows \(g(x) = 8 - 3x\). At the input \(x\), an orange bar shows the height of \(f\), and a green bar on top of it shows the height of \(g\). The top of the stack lies on the dashed curve, which shows the sum \(f + g\).

Move the slider to bring x toward 2 from either side. The height of f approaches 6, and the height of g approaches 2.

At \(x = 1.9\) the heights are 5.61 and 2.3, so the stack is 7.91 high. As x comes closer to 2, the stack comes closer to 8, which is 6 + 2.

Any two functions

As \(x\) approaches the input \(a\), the height of \(f\) approaches a value \(\cy{L}\), and the height of \(g\) approaches a value \(\cy{M}\). The dashed curve shows \(f + g\), and dashed orange lines mark the heights.

Move x toward a. The red pieces show how far each height is from its limit, and they become shorter.

The red piece on the sum is as long as the other two red pieces together, so it becomes shorter when they do.

This means the total approaches \(\cy{L} + \cy{M}\), so the limit of a sum is the sum of the limits. A constant multiple works the same way, because a fixed multiple of a short piece is still short.

Notation

1. Suppose \(f(x)\) approaches \(\cy{L}\) and \(g(x)\) approaches \(\cy{M}\) as \(x\) approaches \(a\). Both limits must exist.

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

2. The limit of a sum is the sum of the limits.

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

   This is the sum law.

3. The limit of a difference is the difference of the limits, in the same order.

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

   This is the difference law.

4. A constant \(k\) comes out in front of the limit. The figure shows \(k = \tfrac12\): the curve and its limit are both half as high.

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

   This is the constant multiple law.

5. Most calculations begin with two basic limits. The limit of a constant is that constant, and the limit of \(x\) as \(x\) approaches \(a\) is \(a\).

   \[\begin{aligned} \lim_{x \to a} k = k \\ \lim_{x \to a} x = a \end{aligned}\]

When it fails

The part of \(g\) to the right of \(a\) is now higher than the part to the left. The heights of \(g\) approach different values from the two sides, so \(g\) has no limit at \(a\).

The laws need a limit for each part, and \(g\) has none, so the sum law does not apply. In this picture the sum also jumps at \(a\), so \(f+g\) has no limit.

Example 1: a linear expression

Find the limit of \(5x - 7\) as \(x\) approaches 4.

1. The solid line is \(5x - 7\). The difference law splits it into \(5x\) and \(7\), shown by the faint dashed lines, and you find the limit of each part.

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

2. The constant multiple law moves the 5 in front. The limit of \(x\) is 4, so the first part approaches 20.

   \[\lim_{x \to 4} 5x = 5 \cdot 4 = \cy{20}\]

3. The limit of the constant 7 is 7.

   \[\lim_{x \to 4} 7 = \cy{7}\]

4. Subtract the two limits. The green segment from 7 up to 20 is 13 long, which is the height the line approaches at \(x = 4\).

   \[20 - 7 = \cy{13}\]

Example 2: a negative limit

As \(x\) approaches 1, \(f(x)\) approaches 5 and \(g(x)\) approaches \(-3\). Find the limit of \(f(x) - 2g(x)\).

1. The difference law and the constant multiple law split the expression. The curve of f approaches 5, and the curve of g approaches −3.

   \[\begin{aligned} &\lim \bigl(f(x) - 2g(x)\bigr) \\ &= \lim f(x) - 2 \lim g(x) \end{aligned}\]

2. A tempting wrong move: Use the sizes of the limits, 5 and 3, and subtract.

   \[5 - 2 \cdot 3 = -1\]

   Because the limit of g is negative, subtracting twice it raises the result above 5. The red dot at −1 lies far below the curve of f, so it cannot be correct.

3. Keep the sign. The limit of \(g\) is \(-3\), so twice it is \(-6\), and subtracting \(-6\) adds 6. The green segment from 5 up to 11 shows this gain.

   \[5 - 2(-3) = 5 + 6 = \cy{11}\]

Practice

As \(x\) approaches 3, \(p(x)\) approaches \(-4\) and \(q(x)\) approaches 10. Find the limit of \(3p(x) + \tfrac12 q(x)\).

The curve of p lies below the x-axis near 3, so its limit is negative. Apply the laws to each term.

Questions

The graph shows \(f\) and \(g\) near \(x = 3\). On \(f\), the open circle marks the height the curve approaches, and the filled dot marks the value \(f(3)\).

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

   • 7
   • 3
   • 10
   Answer

   7

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

   • 1
   • −1
   • 11
   Answer

   1

3. What is the limit of \(9 - 2x\) as \(x\) approaches 5?

   • −1
   • 19
   • −10
   • 1
   Answer

   −1

Summary

You find the limit of a sum, a difference or a constant multiple by combining the separate limits in the same way.

This works because each part stays close to its own limit, so the combination stays close to the combined limits.

Use it when every part has a limit that you know or can find.

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