Pass a limit through a continuous function

A curve with no break

The curve shows \(y = \sqrt{3x+4}\). The orange dot marks the height of the curve at the input \(x\), and the dashed line ends in a tick that shows this height.

Move x toward 4 from either side. At \(x = 3.9\) the height is about 3.96, and at \(x = 4.1\) it is about 4.04, so the heights approach 4.

The curve has no hole and no jump at \(x = 4\), and the filled dot shows the value there, \(\sqrt{16} = 4\). The limit and the value are the same number because the curve is unbroken at that input.

Any unbroken curve

The input \(x\) now approaches a number \(c\), and the orange dot on the curve approaches a height.

A function is continuous at \(c\) if its curve has no hole and no jump at \(c\). This means that \(f(c)\) has a value and that the curve leads to it.

Move x toward \(c\). The heights lead to the filled dot at height \(\cy{f(c)}\), where the orange dashed line ends.

The limit is the height at \(c\), so for a continuous function you find the limit by substituting \(c\).

Notation

1. If \(f\) is continuous at \(c\), the limit as \(x\) approaches \(c\) is the value at \(c\).

   \[\lim_{x \to c} f(x) = \cy{f(c)}\]

   This is direct substitution.

2. Polynomials, roots, \(e^x\), \(\ln x\), \(\sin x\) and \(\cos x\) are continuous at every input inside their domains. Sums, products and compositions of continuous functions are continuous too.

3. For a composition \(f\bigl(g(x)\bigr)\), the limit passes through the outside function \(f\). Suppose \(g(x)\) approaches \(\cy{b}\) as \(x\) approaches \(c\), and \(f\) is continuous at \(\cy{b}\). The faint curve shows \(g\).

   \[\begin{aligned} \lim_{x \to c} f\bigl(g(x)\bigr) &= f\Bigl(\lim_{x \to c} g(x)\Bigr) \\ &= f(\cy{b}) \end{aligned}\]

   This is the limit of a composition.

4. Before you substitute, confirm that the function is defined at the input and on both sides of it. A square root needs a nonnegative inside, and a logarithm needs a positive inside.

When it fails

The curve now starts to the right of \(c\). There is no curve near \(c\), so no heights lie near \(c\) and there is no limit to find. A square root behaves like this where its inside is negative, because the root is not a real number.

Substitution gives no value either, because \(f(c)\) does not exist. The rule needs the function to be defined at the input and on both sides of it.

Example 1: a logarithm

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

1. The inside function \(3x-5\) is continuous, and the faint dashed line shows it approaching 1.

   \[\lim_{x \to 2} (3x-5) = \cy{1}\]

2. The logarithm is continuous at 1, because 1 is positive. The input 2 lies inside the domain, since \(3x-5>0\) when \(x>\tfrac53\), so the limit passes through the logarithm.

   \[\lim_{x \to 2} \ln(3x-5) = \ln 1\]

3. Evaluate. The curve crosses the x-axis at x = 2.

   \[\ln 1 = \cy{0}\]

Example 2: a root of a sum

Find the limit of \(\sqrt{x^2+9}\) as \(x\) approaches 4.

1. The inside function \(x^2+9\) is continuous, so you substitute 4.

   \[\lim_{x \to 4} (x^2+9) = 4^2+9 = 25\]

2. A tempting wrong move: Pass the root onto each term and substitute 4.

   \[\sqrt{4^2} + \sqrt{9} = 4 + 3 = 7\]

   The root applies to the whole inside, not to each term. The red dot at height 7 lies above the curve, which passes through height 5 at x = 4.

3. The root is continuous at 25, so the limit passes through the root of the whole inside.

   \[\sqrt{25} = \cy{5}\]

Practice

Find the limit of \(\sin(2x)\) as \(x\) approaches \(\tfrac{\pi}{6}\).

Find where the inside function is heading. Then evaluate the sine there, because the sine is continuous.

Questions

The graph shows \(f\). At \(x = 3\) the open circle marks the height the curve approaches, and the filled dot marks \(f(3)\). At \(x = 1\) the curve has no break.

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

   • 2
   • 4
   • It does not exist
   Answer

   2

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

   • 4
   • 1
   • It does not exist
   Answer

   4

3. What is the limit of \(\ln(x^3-7)\) as \(x\) approaches 2?

   • 0
   • 1
   • \(\ln 8 - \ln 7\)
   • It does not exist
   Answer

   0

Summary

The limit of a continuous function is its value at the input, and for a composition you take the limit of the inside first.

This works because a continuous function has no break, so the heights near the input lead to the height at the input.

Use it after you confirm that the function is defined at the input and on both sides of it.

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