Limits and Continuity: Intuition First

Consider a function that has a gap in it:

At $x=3$ this is $0/0$, which is not a number, so $f(3)$ does not exist. There is a hole in the graph. But look at what happens as you walk towards the hole from either side:

The outputs are converging on 6 from both directions, and they do it as precisely as you like: get $x$ within a millionth of 3 and $f(x)$ lands within a millionth of 6. Nothing about that behaviour depends on $f(3)$ existing. That approach value is what we call the limit:

The reason it works is visible in the algebra. For every $x$ other than 3, the fraction cancels to $x+3$. The functions $f(x)$ and $x+3$ are not the same function — one has a hole and one does not — but they agree at every point on the approach, and the limit only ever looks at the approach.

This is not a technicality. It is the entire reason calculus works. A derivative is defined as ${\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$, and at $h=0$ that quotient is always $0/0$. If limits required the function to be defined at the point, there would be no derivatives at all.

You may have seen a definition involving $\epsilon$ and $\delta$. That definition is a careful way of saying “as precisely as you like” — for any tolerance you name around the limit, there is a distance around the point that guarantees you stay inside it. It is the right tool for proving a limit. It is almost never the tool for finding one.

Drag those sliders as far as they go. The curve stretches, bends, and flips, but it never develops a hole, a jump, or a vertical asymptote. Polynomials are continuous everywhere, which is why the limit of a polynomial is always found by plain substitution — and why the interesting limits all involve division, roots, piecewise definitions, or infinity.

You need limit techniques exactly when direct substitution fails or is not available. Four situations cover almost everything you will meet.

Substitution gives 0/0. The signature case. The numerator and denominator share a factor that vanishes at the point, and your job is to find and cancel it. Recognising the factorisation quickly is a factoring skill rather than a calculus one, which is why this topic exposes gaps in algebra so ruthlessly.

Substitution gives a non-zero number over zero. There is a vertical asymptote. The two-sided limit does not exist as a number, but the one-sided limits are usually $+\infty$ or $-\infty$, and the sign depends on which side you approach from. Say which.

The variable is heading to infinity. There is nothing to substitute. Compare growth rates instead: divide numerator and denominator by the highest power in the denominator, and every term with a power of $x$ underneath it goes to zero. So ${\lim }_{x\to \infty }\frac{3x^2-x}{5x^2+2}=\frac{3}{5}$, which is the ratio of the leading coefficients.

The point is a boundary of a piecewise function. Different rules apply on each side, so the one-sided limits must be computed separately from their own formulas and then compared.

There is also a small set of special limits worth knowing outright, because no amount of factoring produces them. ${\lim }_{x\to 0}\frac{\sin x}{x}=1$ is the one you will use most, and it generalises: $\frac{\sin (5x)}{3x}$ can be written as $\frac{5}{3}\cdot \frac{\sin (5x)}{5x}$, so its limit is $5/3$.

1. Substitute. Always substitute first. Most limits in a first course are ordinary substitutions dressed up to look frightening, and trying it costs one line. If a number comes out, that is the limit — this is legitimate precisely because polynomials, rational functions, roots, exponentials, and trigonometric functions are continuous everywhere they are defined.

2. Classify the form you landed on. A number means you are done. A non-zero number over zero means an asymptote. Zero over zero, or infinity over infinity, means indeterminate — more work required, and the form tells you which work.

3. Apply the matching algebra. Rational function giving 0/0: factor both parts and cancel the common factor. A square root in a difference: multiply top and bottom by the conjugate, which turns the difference of roots into a difference of squares. A fraction inside a fraction: combine over a common denominator first. Heading to infinity: divide through by the dominant power.

4. Substitute into the simplified expression. This step is valid because cancelling changed the function only at the single point you are approaching, and the limit never looks at that point.

5. Check continuity properly, if that is the question. Three conditions, all required, at the point $x=a$:

Condition (ii) is itself a demand that the two one-sided limits agree. Failing (i) or (iii) gives a removable discontinuity — a hole you could patch by redefining one value. Failing (ii) gives a jump, if the sides disagree by a finite amount, or an infinite discontinuity if the function blows up.

Type a limit and you will get the substitution attempt, the form it produces, and the technique that form calls for — in that order, which is the order you should be working in yourself. The limit calculator with steps handles indeterminate forms, one-sided limits, and limits at infinity, and the calculus solver takes a whole multi-part question at once.

Reading 0/0 as an answer

Writing “the limit is 0”, or “the limit is 1”, or “does not exist” the moment $0/0$ appears. The form is called indeterminate because it genuinely determines nothing: $\frac{x^2}{x}$ gives $0/0$ at zero with limit 0, $\frac{x}{x}$ gives $0/0$ with limit 1, and $\frac{x}{x^3}$ gives $0/0$ with no finite limit at all. Treat it as an instruction to keep working.

Assuming the limit is the function value

Substitution works so often that it starts to feel like the definition, and then a piecewise function or a hole catches you out. The limit equals the function value only when the function is continuous there, which is the third continuity condition, not a general fact. Whenever the point sits at a boundary, at a hole, or where a denominator vanishes, stop and check both sides.

Claiming the cancelled function is the original

After cancelling, people write $\frac{x^2-9}{x-3}=x+3$ without qualification. Those two functions have different domains: one is undefined at 3 and the other is perfectly happy there. The correct statement adds $x\ne 3$. It matters when a question asks you to describe the graph or to classify the discontinuity, because the hole is the whole point.

Ignoring the two sides at an asymptote

${\lim }_{x\to 0}\frac{1}{x}$ does not exist, because approaching from the right sends the function to $+\infty$ and approaching from the left sends it to $-\infty$. Writing $\infty$ as the answer hides the disagreement. The same trap appears with $\frac{|x|}{x}$, which is $+1$ on one side and $-1$ on the other, so the limit fails to exist even though nothing blows up. If the algebra ever produces a non-zero number over zero, ask about the sign on each side before you write anything down.

One of each of the four situations above. Give exact values, not decimals.