Two-Sided Limits

Example With $a = 2$ and $\delta = 0.5$ the punctured neighborhood is $(1.5, 2) \cup (2, 2.5)$. It contains $1.7$ and $2.4999$ but not $2$ and not $2.5$.

Example The function

is undefined at $2$, but for $x \neq 2$ it simplifies to

So on every punctured neighborhood of $2$ the function equals $x + 3$, and $\lim_{x \to 2} f(x) = 5$.

Example Let $f(x) = x^2$ for $x < 2$ and $f(x) = 2x$ for $x > 2$, with $f(2)$ left undefined. The left-hand limit is $2^2 = 4$ and the right-hand limit is $2 \cdot 2 = 4$. They agree, so $\lim_{x \to 2} f(x) = 4$. If instead the right piece were $2x + 1$, the right-hand limit would be $5$ and the two-sided limit would not exist.

Example Uniqueness gives a quick test for nonexistence. For $\operatorname{sgn}(x)$, which is $-1$ for $x < 0$, $0$ at $0$ and $1$ for $x > 0$, the inputs $x = 1/n$ for positive whole numbers $n$ approach $0$ and give the output $1$, while the inputs $x = -1/n$ also approach $0$ and give $-1$. A limit $L$ would have to be close to both $1$ and $-1$, so $\lim_{x \to 0} \operatorname{sgn}(x)$ does not exist.

Example The function $\sqrt{x}$ has domain $[0, \infty)$, so at $0$ only the right side is available, and $\lim_{x \to 0^+} \sqrt{x} = 0$. The two-sided limit $\lim_{x \to 0} \sqrt{x}$ is not defined, since $\sqrt{x}$ has no values for $x < 0$.

A side note. Some advanced texts use a looser convention in which a two-sided limit approaches $a$ only through points of the domain. In that convention $\lim_{x \to 0} \sqrt{x} = 0$. This course does not use it, except where the formal definition mentions it.