Limits · Definition

Limit of a function

calculuslimits

Before you start you need to know

Before computing limits, it helps to understand what it means to approach a value without necessarily reaching it, and why \(f(a)\) may not matter.

Definition

The limit of \(f\) as \(x\) approaches \(a\) is \(L\) if \(f(x)\) gets as close to \(L\) as we like when \(x\) is taken sufficiently close to \(a\), regardless of what happens at \(x=a\).

\[\lim_{x\to a} f(x)=L\]

A limit answers one very specific question: "where is \(f(x)\) heading as \(x\) gets close to \(a\)?" Notice that the question is not "what is \(f(a)\)?"; the function does not even need to be defined at \(a\).

A classic example is \(\frac{x^2-1}{x-1}\) at \(x=1\): the expression does not exist there, but for any nearby \(x\) (other than 1) it equals \(x+1\), so the values approach 2. The limit is 2 even though \(f(1)\) does not exist.

The formal (epsilon–delta) definition just makes "as close as we like" precise: for every output tolerance \(\varepsilon>0\) there is an input distance \(\delta>0\) that guarantees it. All of calculus rests on this idea.

Think about it

Can the limit at a exist if f(a) is not defined?

Yes. The limit only looks at values near a, not at the value at a.

Using the definition, prove that \(\lim_{x\to2}x^2=4\).

If \(|x-2|<\delta\le1\), then \(|x+2|<5\) and \(|x^2-4|=|x-2||x+2|<5\delta\). Taking \(\delta=\min\{1,\varepsilon/5\}\) gives \(|x^2-4|<\varepsilon\).

Exercises 20

References

  • James Stewart, Calculus: Early Transcendentals, 8.ª edición, Cengage Learning, 2015.
  • Michael Spivak, Calculus, 4.ª edición, Publish or Perish, 2008.
  • Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §4.2, Definición 4.2.1
  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 4, Definición 4.1
  • Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §4.1, Definición 4.1.4