Limits of Functions

In this lesson you’ll learn the epsilon-delta definition of a function limit, the standard proof template, why the point $a$ itself is excluded, and the sequential criterion that makes disproving limits easy. This is the definition the whole subject is built on.

$\displaystyle\lim_{x\to a} f(x) = L$ means

$$\forall \varepsilon>0 \; \exists \delta>0 \; \text{such that} \; 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon$$

The same game as epsilon-N. Your opponent picks a tolerance $\varepsilon$ on the output; you must produce a tolerance $\delta$ on the input that guarantees it.

Geometrically: draw a horizontal band of half-width $\varepsilon$ around $L$. You must find a vertical strip of half-width $\delta$ around $a$ such that the graph over that strip stays inside the band. Thinner band, thinner strip.

That strict inequality excludes $x = a$, and it is deliberate.

The limit does not care about $f(a)$. The function need not even be defined at $a$. This is essential, because the derivative is defined as

$$\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$$

and that quotient is undefined at $h = 0$. If limits required the value at the point, calculus could not start.

So three situations are all consistent with $\lim_{x\to a} f(x) = L$: $f(a) = L$, $f(a) \ne L$, and $f(a)$ undefined. Continuity is the extra condition that $f(a)$ exists and equals $L$ - the next lesson.

Same shape as before.

1. Scratch. Start from $|f(x)-L| < \varepsilon$. Factor out $|x-a|$ and bound whatever is left over.
2. Choose $\delta$. Usually a min of two things.
3. Verify. Assume $0<|x-a|<\delta$, derive the conclusion forwards.

For a linear function it’s clean. To show $\lim_{x\to 3}(2x+1) = 7$:

Scratch. $|(2x+1) - 7| = |2x - 6| = 2|x-3|$, and $2|x-3| < \varepsilon \iff |x-3| < \frac{\varepsilon}{2}$.

Proof. Given $\varepsilon>0$, take $\delta = \frac{\varepsilon}{2}$. If $0<|x-3|<\delta$ then $|(2x+1)-7| = 2|x-3| < 2\delta = \varepsilon$. ∎

Non-linear functions need one extra step. For $\lim_{x\to 2} x^2 = 4$:

$$|x^2 - 4| = |x-2||x+2|$$

We control $|x-2|$, but $|x+2|$ is a nuisance factor. Bound it first by restricting $\delta$.

If $\delta \le 1$ then $|x-2|<1$ gives $1 < x < 3$, so $|x+2| < 5$. Then

$$|x^2-4| < 5|x-2|$$

which is below $\varepsilon$ once $|x-2| < \frac{\varepsilon}{5}$. So take

$$\delta = \min\left(1, \frac{\varepsilon}{5}\right)$$

The min does two jobs at once: the 1 tames the nuisance factor, and the $\frac{\varepsilon}{5}$ delivers the tolerance. This pattern appears in nearly every non-linear epsilon-delta proof.

$\displaystyle\lim_{x\to a} f(x) = L$ $\iff$ for every sequence $x_n \to a$ with $x_n \ne a$, we have $f(x_n) \to L$.

This connects function limits to sequence limits, and it hands you two useful things.

All the sequence limit theorems transfer. Sums, products, quotients and the squeeze theorem for function limits follow immediately, with no new epsilon-delta work.

Disproving a limit becomes easy. Exhibit two sequences approaching $a$ whose images have different limits, and the limit cannot exist. Negating the epsilon-delta definition directly is much more painful.

$\lim_{x\to a^+} f(x) = L$ uses $a < x < a+\delta$; the left limit uses $a - \delta < x < a$. Then

$$\lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L$$

For $f(x) = \frac{|x|}{x}$ the right limit at 0 is 1 and the left is $-1$, so the two-sided limit does not exist.

Example 1: Prove $\lim_{x\to 4}(3x - 5) = 7$.

Solution. Scratch. $|(3x-5)-7| = |3x-12| = 3|x-4| < \varepsilon \iff |x-4| < \frac{\varepsilon}{3}$.

Proof. Let $\varepsilon>0$ and take $\delta = \frac{\varepsilon}{3}$. If $0<|x-4|<\delta$ then

$$|(3x-5)-7| = 3|x-4| < 3\delta = \varepsilon$$

∎

For any linear $f(x) = mx+c$ with $m \ne 0$, $\delta = \frac{\varepsilon}{|m|}$ always works.

Example 2: Prove $\lim_{x\to 3} x^2 = 9$.

Solution. Scratch. $|x^2-9| = |x-3||x+3|$. Restrict $\delta \le 1$, so $2<x<4$ and $|x+3| < 7$. Then $|x^2-9| < 7|x-3|$, below $\varepsilon$ when $|x-3| < \frac{\varepsilon}{7}$.

Proof. Given $\varepsilon>0$, take $\delta = \min\left(1, \frac{\varepsilon}{7}\right)$. If $0<|x-3|<\delta$ then $|x-3|<1$ so $|x+3|<7$, and

$$|x^2-9| = |x-3||x+3| < 7|x-3| < 7 \cdot \frac{\varepsilon}{7} = \varepsilon$$

∎

Example 3: A limit where the function is undefined at the point.

Compute $\lim_{x\to 2}\frac{x^2-4}{x-2}$.

Solution. At $x = 2$ the expression is $\frac00$, undefined. But the definition only looks at $x \ne 2$, and there

$$\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2$$

The cancellation is legal precisely because $x \ne 2$. So the limit is $\lim_{x\to2}(x+2) = 4$.

The limit is 4 even though $f(2)$ does not exist. This is exactly the situation every derivative is in.

Example 4: Disprove a limit with sequences.

Show $\lim_{x\to 0}\sin\left(\frac1x\right)$ does not exist.

Solution. Take two sequences approaching 0.

$$x_n = \frac{1}{2\pi n} \implies \sin\left(\frac{1}{x_n}\right) = \sin(2\pi n) = 0$$ $$y_n = \frac{1}{2\pi n + \frac{\pi}{2}} \implies \sin\left(\frac{1}{y_n}\right) = \sin\left(2\pi n + \frac\pi2\right) = 1$$

Both $x_n \to 0$ and $y_n \to 0$, but the images converge to 0 and 1.

By the sequential criterion, the limit does not exist. ∎

Compare how much work negating the epsilon-delta definition would have been. Two sequences settled it.

Example 5: Delta depending on the point.

For $f(x) = \frac1x$ on $(0,\infty)$, how does $\delta$ depend on $a$ for a fixed $\varepsilon$?

Solution.

$$\left|\frac1x - \frac1a\right| = \frac{|a-x|}{|ax|}$$

Restrict $|x - a| < \frac a2$, so $x > \frac a2$ and $|ax| > \frac{a^2}{2}$. Then

$$\left|\frac1x-\frac1a\right| < \frac{2|x-a|}{a^2}$$

so $\delta = \min\left(\frac a2, \frac{\varepsilon a^2}{2}\right)$ works.

$\delta$ shrinks like $a^2$ as $a \to 0$. The function gets steeper near 0, so the same output tolerance demands a much tighter input tolerance. That dependence on the point, not just on $\varepsilon$, is what fails to be uniform - and uniform continuity, four lessons ahead, is exactly the condition that rules it out.

Example 6: A limit of 0 via squeezing.

Compute $\lim_{x\to 0} x\sin\left(\frac1x\right)$.

Solution. The previous example shows $\sin\frac1x$ has no limit. But it is bounded, and

$$\left|x\sin\frac1x\right| \le |x|$$

so $-|x| \le x\sin\frac1x \le |x|$. Both bounds $\to 0$, so by squeezing the limit is 0.

A bounded factor times a vanishing one goes to zero, whether or not the bounded factor has a limit. Same result as for sequences, now for functions.

Example 7: One-sided limits differing.

Analyse $f(x) = \frac{|x|}{x}$ at $x = 0$.

Solution. For $x>0$, $f(x) = 1$. For $x<0$, $f(x) = -1$. Undefined at 0.

$$\lim_{x\to0^+}f(x) = 1, \qquad \lim_{x\to0^-}f(x) = -1$$

Different, so the two-sided limit does not exist. Sequentially: $\frac1n \to 0$ gives images 1, and $-\frac1n \to 0$ gives images $-1$.

This is a jump discontinuity, and it is what makes the sign function non-integrable-by-inspection arguments fail and why step functions need care.

Tolerance engineering. “The output must be within $\varepsilon$, so how tightly must the input be controlled?” is literally the epsilon-delta question. Manufacturing tolerance stack-up analysis computes $\delta$ from a required $\varepsilon$, and the derivative is the local conversion factor between them.

Numerical differentiation. Approximating $f'(a)$ by a difference quotient with small $h$ relies on the limit existing. Too large an $h$ gives truncation error and too small an $h$ gives catastrophic floating-point cancellation, so there is an optimal $h$ - a very practical consequence of the limit being approached rather than reached.

Sensor calibration. A sensor’s sensitivity is a derivative, and the amplification of input noise into output noise is the $\frac{\varepsilon}{\delta}$ ratio. Example 5’s point matters here: near a steep part of the response curve, the same output precision demands far better input precision.

Control systems. Steady-state error is a limit as $t \to \infty$, and a controller specification like “settle within 2% of setpoint” is a choice of $\varepsilon$ with a corresponding settling time.

Graphics and continuity of rendering. A shader with a discontinuity produces visible seams. Detecting where a piecewise definition fails to have matching one-sided limits is exactly how those artifacts get diagnosed.