Two-Sided and One-Sided Limits

The equivalence below assumes domain points approach $a$ from both sides. At an endpoint such as $0$ for $\sqrt{x}$, use the right-hand limit; a missing left-hand domain is not a jump. A limit relative to a domain, as in the definition, need not be a two-sided limit.

A finite two-sided limit exists exactly when both one-sided limits exist and agree:

$$\lim_{x\to a}f(x)=L \quad\Longleftrightarrow\quad \lim_{x\to a^-}f(x)=L \ \text{and}\ \lim_{x\to a^+}f(x)=L.$$

This gives a reliable test at a piecewise boundary:

1. evaluate the expression used for $x<a$ to obtain the left-hand limit;
2. evaluate the expression used for $x>a$ to obtain the right-hand limit;
3. compare the results.

If the two values differ, the two-sided limit does not exist. For example,

$$f(x)= \begin{cases} 0, & x<0,\\ 1, & x\ge 0 \end{cases}$$

has left-hand limit $0$ and right-hand limit $1$ at the origin, so $\lim_{x\to0}f(x)$ does not exist.

The value $f(a)$ is a separate question. A two-sided limit can exist when $f(a)$ is missing or differs from the limit; equality with $f(a)$ is required for continuity, not for the limit itself.

Nonexistence can also come from oscillation, not just unequal finite limits. For $f(x)=\sin(1/x)$, the positive sequences $x_n=1/(\pi/2+2\pi n)$ and $y_n=1/(3\pi/2+2\pi n)$ both approach $0$, but $f(x_n)=1$ and $f(y_n)=-1$. Even the right-hand limit fails. By contrast, $1/x^2\to+\infty$ from both sides, while $1/x$ tends to $-\infty$ from the left and $+\infty$ from the right.

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