Limits · Definition

Indeterminate forms

calculuslimits

Before you start you need to know

They appear when computing limits and direct substitution gives an expression with no defined value. To resolve them, transform the expression before deciding the result.

Definition

An indeterminate form is a limit expression, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), whose value cannot be deduced from the limits of its parts.

\[\frac{0}{0},\ \frac{\infty}{\infty},\ 0\cdot\infty,\ \infty-\infty\]

An indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) is not a result: it is a warning that direct substitution is not enough. Depending on how numerator and denominator "compete", the limit can be any number, infinite, or nonexistent.

Typical mistake: concluding that \(\frac{0}{0}\) equals 0 or 1 without analyzing the expression. The same appearance can give different limits, so transform before deciding.

Example: \(\frac{x^2-9}{x-3}\) gives \(\frac{0}{0}\) at \(x=3\), but it factors as \(x+3\) and the limit is 6.

Think about it

Is 0/0 equal to 1 or to 0?

Neither: it is indeterminate; it depends on the specific functions and you must rewrite the expression.

Exercises 10

Compute the following limits

  1. \(\displaystyle \lim_{x\to 3}\,\frac{x^{2} - 9}{x - 3}\)
  2. \(\displaystyle \lim_{x\to 2}\,\frac{x^{2} - 5 x + 6}{x - 2}\)
  3. \(\displaystyle \lim_{x\to 2}\,\frac{x^{3} - 8}{x - 2}\)
  4. \(\displaystyle \lim_{x\to 2}\,\frac{x^{2} + x - 6}{x^{2} - 4}\)
  5. \(\displaystyle \lim_{x\to 1}\,\frac{x^{2} - 1}{x^{2} - 3 x + 2}\)
  6. \(\displaystyle \lim_{x\to 0}\,\frac{\sqrt{x + 1} - 1}{x}\)
  7. \(\displaystyle \lim_{x\to 0}\,\frac{\sqrt{x + 4} - 2}{x}\)

Problems

  1. Which of these limits has the indeterminate form 0/0 on direct substitution?
  2. Which of these expressions is an indeterminate form?
  3. Which of these expressions has the form \(\infty-\infty\)?

What comes next

References

  • James Stewart, Calculus: Early Transcendentals, 8.ª edición, Cengage Learning, 2015.