Introduction to limits | Calculus

Getting close without arriving

The function $f(x) = \frac{x^2 - 1}{x - 1}$ isn't defined at $$x = 1$$ . But what happens near $$1$$ ?

$$x$$	$$0.9$$	$$0.99$$	$$0.999$$	$$1.001$$	$$1.01$$	$$1.1$$
$$f(x)$$	$$1.9$$	$$1.99$$	$$1.999$$	$$2.001$$	$$2.01$$	$$2.1$$

As $$x$$ gets close to $$1$$ , $$f(x)$$ gets close to $$2$$ . We write

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2

(For $x \neq 1$ , the function simplifies to $$x + 1$$ .)

The value $$f(c)$$ itself doesn't matter. It may be undefined, or different from the limit. Only the behavior near $$c$$ counts.

One-sided limits

Infinite limits

Near $$x = 0$$ , $\frac{1}{x^2}$ gets huge: $\frac{1}{0.001^2} = 1\,000\,000$ . We write

\lim_{x \to 0} \frac{1}{x^2} = \infty

Strictly speaking the limit doesn't exist (no finite value), but writing $\infty$ describes how it fails: the function grows without bound. The line $$x = 0$$ is a vertical asymptote.

Limits at infinity

We can also ask what happens as $$x$$ grows without bound:

\lim_{x \to \infty} \frac{1}{x} = 0 \qquad\qquad \lim_{x \to -\infty} e^x = 0 \qquad\qquad \lim_{x \to \infty} \sqrt{x} = \infty

A finite limit at $\pm\infty$ gives a horizontal asymptote.

Three ways a limit can fail to exist

The one-sided limits are different (a jump).
The function grows without bound (a vertical asymptote).
The function oscillates without settling, like $\sin\frac{1}{x}$ near $$0$$ .

Practice

Quick check

If $\lim_{x \to 3^-} f(x) = 7$ and $\lim_{x \to 3^+} f(x) = 7$ , but $$f(3) = 1$$ , what is $\lim_{x \to 3} f(x)$ ?

Quick check

What is $\lim_{x \to \infty} \frac{1}{x^2}$ ?