Epsilon-Delta Definition of a Limit

A function might approach a value as the input moves toward a point. The outputs get closer and closer to some number, and inputs closer to the point give outputs closer to that number. At some stage, you want to say the function has a limit there. But that statement is imprecise until you say what closeness means and how much of it you require. The epsilon-delta definition settles this. It turns the informal idea of approaching into a precise condition relating input proximity to output proximity.

The statement $\lim_{x \to a} f(x) = L$ means that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that

\[0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.\]

The number $\varepsilon$ sets the tolerance on the output. It specifies how close to $L$ you require $f(x)$ to be. The number $\delta$ is chosen depending on $\varepsilon$. It specifies how close to $a$ the input must be so that the tolerance is met. The definition requires this to hold for every positive $\varepsilon$, no matter how small. You can shrink $\varepsilon$ as far as you like, and a suitable $\delta$ must still exist. The condition $0 < \lvert x - a \rvert$ excludes $x = a$ itself. The limit concerns the behavior of the function near the point, not at the point.

How do you prove a limit with this definition? Start with an arbitrary $\varepsilon > 0$ and work backward. Begin with the inequality you need, $\lvert f(x) - L \rvert < \varepsilon$, and manipulate it until it becomes a condition on $\lvert x - a \rvert$. That condition determines the choice of $\delta$. For a linear function, $\lvert f(x) - L \rvert$ is proportional to $\lvert x - a \rvert$, and the choice is immediate. For nonlinear functions, the relationship involves extra factors that depend on $x$. You control these by first restricting $\delta$ to a preliminary bound that keeps $x$ in a bounded interval, then choose the final $\delta$ to satisfy both the restriction and the $\varepsilon$ requirement.

Take a concrete case. Prove that $\lim_{x \to 2} x^2 = 4$. The distance between $f(x) = x^2$ and the proposed limit is $\lvert x^2 - 4 \rvert = \lvert x - 2 \rvert\,\lvert x + 2 \rvert$. The factor $\lvert x - 2 \rvert$ is the input proximity you can make small. The factor $\lvert x + 2 \rvert$ depends on $x$ and must be bounded before you can isolate $\lvert x - 2 \rvert$. Impose the preliminary restriction $\delta \leq 1$. Then $\lvert x - 2 \rvert < 1$ forces $x$ into the interval $(1, 3)$, which makes $\lvert x + 2 \rvert$ less than $5$. Under this restriction, $\lvert x^2 - 4 \rvert < 5\,\lvert x - 2 \rvert$. You need $5\,\lvert x - 2 \rvert < \varepsilon$, and this holds whenever $\lvert x - 2 \rvert < \varepsilon / 5$. Set $\delta = \min(1,\, \varepsilon / 5)$. If $\varepsilon = 0.1$, then $\delta = 0.02$, and any $x$ within $0.02$ of $2$ satisfies $\lvert x^2 - 4 \rvert < 0.1$. If $\varepsilon = 0.001$, then $\delta = 0.0002$, and the bound still holds. No matter how small $\varepsilon$ gets, this formula produces a valid $\delta$.

The definition says nothing about $f(a)$. The function does not even need to be defined at $a$. The expression $(x^2 - 4)/(x - 2)$ is undefined at $x = 2$, but for every other $x$ it simplifies to $x + 2$. The epsilon-delta definition confirms that the limit as $x \to 2$ is $4$, because closeness of $x$ to $2$ forces closeness of the output to $4$ regardless of what happens at $2$ itself. This separation of the limit from the function value is exactly what makes limits useful. They describe the behavior of a function near a point even when the function is not well-behaved at the point. Continuity is the special case where the limit and the value agree, meaning $\lim_{x \to a} f(x) = f(a)$.