Introduction to Limits
Define the limit intuitively: lim(x→a) f(x) = L means f(x) approaches L as x approaches a from both sides. Estimate limits from graphs and tables of values, and understand that the limit is independent of f(a).
A lesson of Calculus, in Limits and Continuity, about 10 minutes.
By the end of this lesson, you will be able to state the formal ($\varepsilon$-$\delta$) definition of a limit, connect it to the intuitive approach, evaluate limits numerically and graphically with precision, and understand why the limit is the foundational concept of calculus.
Why Limits Matter
Calculus could be called "the mathematics of infinity and infinitesimals." Every central object — the derivative, the definite integral, infinite series — is defined as a limit. The limit isn't a calculator shortcut; it's the logical foundation that makes calculus rigorous.
The Formal Definition ($\varepsilon$-$\delta$)
$$\lim_{x \to c} f(x) = L$$
means: for every $\varepsilon > 0$, there exists $\delta > 0$ such that if $0 < |x - c| < \delta$, then $|f(x) - L| < \varepsilon$.
Translation: No matter how small you want the error in $f(x)$ to be (that's $\varepsilon$), we can always find inputs within $\delta$ of $c$ that achieve that precision. The function can be made arbitrarily close to $L$ by choosing $x$ close enough to $c$.
The condition $0 < |x - c|$ means $x \neq c$ — the limit doesn't care about $f(c)$ itself.
Intuitive vs. Formal
Intuitive: "$f(x)$ gets closer to $L$ as $x$ gets closer to $c$."
Formal: specifies how close to $c$ ensures how close to $L$. The $\varepsilon$-$\delta$ definition eliminates all ambiguity.
Why bother? Without the formal definition, we can't prove that $\lim_{x\to 0}\sin x/x = 1$ rigorously — we can only "observe" it numerically.
Example 1 — A Formal Limit Proof
Prove: $\lim_{x \to 3}(2x - 1) = 5$.
Given $\varepsilon > 0$, we need $\delta$ such that $|x - 3| < \delta \implies |(2x-1) - 5| < \varepsilon$.
$|(2x-1) - 5| = |2x - 6| = 2|x - 3| < \varepsilon$
So we need $2|x-3| < \varepsilon$, i.e., $|x-3| < \varepsilon/2$.
Choose $\delta = \varepsilon/2$.
Then if $|x-3| < \delta = \varepsilon/2$: $|f(x) - L| = 2|x-3| < 2(\varepsilon/2) = \varepsilon$. ✓ ∎
Example 2 — The Limit at a Removable Discontinuity
$f(x) = \dfrac{x^2-4}{x-2}$ at $x = 2$:
$f(2)$ is undefined (0/0). But:
$\lim_{x \to 2}f(x) = \lim_{x \to 2}(x+2) = 4$
Formally: $|(x+2) - 4| = |x-2| < \varepsilon$ whenever $|x-2| < \varepsilon$. Choose $\delta = \varepsilon$.
Important Limit Results (No Proof Here)
$$\lim_{x \to 0}\frac{\sin x}{x} = 1 \qquad \lim_{x \to 0}\frac{1-\cos x}{x} = 0 \qquad \lim_{x \to 0}(1+x)^{1/x} = e$$