A limit asks where you are heading, not where you land
Consider f(x) = (x² − 1)/(x − 1) at x = 1. Substituting gives 0/0, which is meaningless. But factorise the top into (x−1)(x+1), cancel, and you are left with x + 1 everywhere except at the one forbidden point.
So as x creeps toward 1 from either side, f(x) creeps toward 2 — even though f(1) does not exist at all. That is the whole idea: a limit describes the destination the function is aiming for, and does not care whether the function actually gets there.
Left and right have to agree
Approach a from below and you get the left-hand limit. Approach from above and you get the right-hand limit. The limit exists only if the two are equal.
For f(x) = |x|/x at zero, the left-hand limit is −1 and the right-hand limit is +1. They disagree, so the limit does not exist — the graph jumps. This is exactly how the sharp corners and jumps in a function get detected.
"Does not exist" is a valid answer
Students often assume every limit must be a number. A limit fails to exist when the two sides disagree, when the function oscillates forever like sin(1/x) near zero, or when it runs off to infinity. Saying so is the correct answer, not a failure to solve.
The laws let you break big limits into small ones
Limits pass straight through addition, subtraction, multiplication and division — with one caveat on division. That means you can evaluate a complicated expression piece by piece instead of all at once.
For a well-behaved function, "taking the limit" is just substitution. The interesting cases are precisely the ones where substitution fails: 0/0 and ∞/∞. Those are indeterminate forms, and they are a signal to factorise, rationalise or cancel before trying again.
| Law | Statement |
|---|---|
| Sum | lim(f + g) = lim f + lim g |
| Product | lim(f · g) = lim f · lim g |
| Quotient | lim(f / g) = lim f / lim g, provided lim g ≠ 0 |
| Power | lim(fⁿ) = (lim f)ⁿ |
| Standard | lim(θ→0) sin θ / θ = 1, with θ in radians |
Continuity is the limit behaving itself
A function is continuous at a when three things all hold: f(a) exists, the limit as x → a exists, and the two are equal. Informally, you can draw it through that point without lifting your pen.
Continuity is what everything later depends on. The derivative is built from a limit, so a function with a jump has no derivative there. This is why limits come first in every calculus course — they are the foundation the rest is poured onto.
Ignore the tangent line for a moment and just watch the gap readout as you drag h toward zero. The secant slope never equals the tangent slope at any h you can set — but it approaches it. That approach is the limit, and it is how the derivative is defined.
When substitution gives 0/0
Most limits are found by simply substituting the value — that is what continuity means. The interesting ones are those where substitution produces 0/0, which is not an answer but a signal: the expression is indeterminate and needs rearranging before it will tell you anything.
Two techniques handle almost every case at this level. If the expression is a ratio of polynomials, factorise and cancel the factor that is causing the zero. If it contains a square root, multiply by the conjugate to move the root out of the way.
Evaluate lim(x→3) (x² − 9)/(x − 3) and lim(x→0) (√(x+4) − 2)/x.
- For the first, substituting gives 0/0, so factorise:
(x − 3)(x + 3)/(x − 3).The common factor (x − 3) is exactly what is making both parts zero. - Cancel — legitimate because x approaches 3 without ever equalling it — leaving
x + 3, so the limit is 6.The function is undefined at x = 3, but the limit does not care what happens at the point itself. - For the second, multiply top and bottom by the conjugate
√(x+4) + 2.The numerator becomes a difference of two squares and the root disappears. - Numerator:
(x + 4) − 4 = x, so the expression isx / [x(√(x+4) + 2)] = 1/(√(x+4) + 2).The x now cancels, which was the whole purpose. - Substituting x = 0 gives
1/(2 + 2) = 1/4.With the indeterminate form removed, ordinary substitution works.
First limit 6; second limit 1/4
Limits at infinity, and the two standard results
Asking what happens as x → ∞ is asking about the long-run behaviour of a function — the shape of its graph far out to the right. For a ratio of polynomials the method is mechanical: divide every term, top and bottom, by the highest power of x appearing in the denominator, then use the fact that 1/xⁿ → 0.
The outcome depends only on the degrees. Equal degrees give the ratio of the leading coefficients; a larger degree on top sends the limit to infinity; a larger degree below sends it to zero.
∞ is not a number you may substitute
Writing ∞/∞ = 1 or ∞ − ∞ = 0 is meaningless, and both are indeterminate forms exactly like 0/0. Divide through by the highest power first and the expression will resolve into ordinary arithmetic. The same applies to 0 × ∞: rearrange it into a quotient before evaluating.