Limits

A limit describes the value that a function approaches as the input approaches a specified number. The notation $\lim_{x\to a}f(x)=L$ means that values of $f(x)$ can be made as close to $L$ as desired by taking $x$ sufficiently close to $a$, without requiring $x=a$. This separation between approach and value explains why a limit may exist at a hole and why knowing $f(a)$ alone does not determine the limit.

Approach must be checked from both directions. The left-hand limit $\lim_{x\to a^-}f(x)$ uses inputs smaller than $a$; the right-hand limit $\lim_{x\to a^+}f(x)$ uses inputs larger than $a$. The two-sided limit exists exactly when both one-sided limits exist and are equal. A jump gives two finite but unequal one-sided limits; a vertical asymptote may produce infinite one-sided behaviour. Neither situation has a finite two-sided limit.

A function is continuous at $x=a$ only when three conditions all hold: $f(a)$ is defined, $\lim_{x\to a}f(x)$ exists, and the limit equals $f(a)$. A removable discontinuity fails the first or third condition while preserving a common approached value. A jump fails the equality of one-sided limits. An infinite discontinuity occurs when function values grow without bound near the point.

Limits can be estimated numerically from a table and visually from a graph, but algebra supplies the exact conclusion. Begin with direct substitution. If it gives an ordinary number in a continuous expression, that number is the limit. If it gives $0/0$, the form is indeterminate: factor and cancel, rationalize, or combine fractions before substituting again. A nonzero number over zero is different and signals unbounded behaviour that must be checked from each side.

The squeeze theorem handles a function trapped between two others. If $g(x)\le f(x)\le h(x)$ near $a$ and both outside functions approach the same number $L$, then $f(x)$ must also approach $L$. For example, $-x^2\le x^2\sin(1/x)\le x^2$ and both bounds approach zero, so $\lim_{x\to0}x^2\sin(1/x)=0$ even though the sine factor oscillates infinitely often.

The word “approaches” can be made exact. The statement $\lim_{x\to a}f(x)=L$ means that for every tolerance $\varepsilon>0$ there is a distance $\delta>0$ such that whenever $0<|x-a|<\delta$, the output satisfies $|f(x)-L|<\varepsilon$. Read plainly: name how close to $L$ you insist the outputs land, and there is always a small interval around $a$ — excluding $a$ itself — on which the function meets that demand. The clause $0<|x-a|$ is the whole reason a limit ignores the value at $a$, because the point $x=a$ is deliberately excluded from the test. You will not write $\varepsilon$–$\delta$ proofs in this course, but this definition is the precise engine that the intuitive phrase “gets arbitrarily close” is standing in for, and it is why numerical tables and graphs can only ever suggest a limit while algebra confirms it.

A limit is a question about the neighbourhood of a point, not the point itself. When we write $\lim_{x\to c} f(x)$ we are asking: as $x$ gets arbitrarily close to $c$ — from both sides, but never actually equal to $c$ — what single number is $f(x)$ closing in on? The value $f(c)$ might not even exist, and that is fine. The limit can exist where the function does not, and the function can exist where the limit does not. Keeping these two questions separate is the single most important habit in this unit.

You will factor quadratics, use the difference of squares $a^2-b^2=(a-b)(a+b)$, rationalise with conjugates, and add fractions over a common denominator. If any of those feels shaky, pause and refresh it — the calculus here is easy once the algebra is automatic.

The fastest way to answer 'where is $f$ heading?' is often the most obvious one: put $x=c$ into the function. This works exactly whenever $f$ is continuous at $c$, because continuity is defined by $\lim_{x\to c} f(x)=f(c)$ — the curve arrives at the point with no jump, no hole and no blow-up.

Which functions are safe to substitute into? Polynomials are continuous everywhere. A rational function is continuous everywhere its denominator is non-zero. A root function is continuous wherever the expression inside is non-negative. So for these families, if plugging in $c$ gives an ordinary defined number, that number is the limit — you are done, and any extra algebra is wasted effort.

When you cannot simply read the answer off, the limit laws let you take a complicated expression apart. If $\lim_{x\to c} f(x)$ and $\lim_{x\to c} g(x)$ both exist and are finite, then the limit of a sum, difference, constant multiple, product, quotient, power or root is assembled from the limits of the parts. Two conditions are worth watching: the quotient law needs the denominator’s limit to be non-zero, and an even root needs the radicand’s limit to be non-negative.

These laws are also what justify substitution in the first place. Because $\lim_{x\to c} x = c$ and $\lim_{x\to c} k = k$ are obviously true, repeatedly applying the sum, product and constant-multiple laws rebuilds any polynomial as $p(c)$. The laws are the engine underneath; substitution is the shortcut you earn from them once you know the function is continuous.

Substitution fails when it produces a shape the laws cannot resolve. The most common is $\tfrac{0}{0}$. This is called an indeterminate form, and the word is exact: $\tfrac{0}{0}$ does not mean the limit is $0$, and it does not mean the limit fails to exist. It means this calculation, by itself, has decided nothing yet — you must do more work before you know the answer.

Here is why 'indeterminate' is the right word. Every one of these expressions gives $\tfrac{0}{0}$ at $x=3$, yet they head to completely different places: $\dfrac{x^2-9}{x-3}\to 6$; $\dfrac{x-3}{x^2-9}\to \tfrac16$; $\dfrac{(x-3)^2}{x-3}\to 0$; and $\dfrac{x-3}{(x-3)^2}$ blows up to $\pm\infty$ and has no limit at all. Same $\tfrac{0}{0}$ reading, four different fates. That is precisely why $\tfrac{0}{0}$ can never be a final answer — it is a signal to simplify, not a verdict.

Geometrically, a $\tfrac{0}{0}$ almost always marks a removable discontinuity — a single-point hole in an otherwise smooth curve. The numerator and denominator share a factor that vanishes at $c$. Everywhere except $c$ that factor is a perfectly ordinary non-zero number you are allowed to divide out, so the function secretly agrees with a simpler expression at every point near the hole. The limit is just the height the simpler curve is reaching for.

Not every 'zero in the denominator' is $\tfrac{0}{0}$. Consider $\lim_{x\to 2}\dfrac{x+1}{x-2}$. At $x=2$ the numerator is $3$ (not zero) while the denominator is $0$, so the form is $\tfrac{3}{0}$ — not indeterminate. There is no shared factor to cancel; instead the fraction grows without bound. Just to the right of $2$ it races to $+\infty$, just to the left it plunges to $-\infty$, so the two-sided limit does not exist and the graph has a vertical asymptote. Diagnosing which case you are in — $\tfrac{0}{0}$ (simplify) versus $\tfrac{\text{non-zero}}{0}$ (asymptote) — is half the skill.

1 · Factor and cancel. When numerator and denominator are polynomials, factor both, cancel the shared $(x-c)$ factor, and substitute into what remains. Difference of squares, common monomial factors and simple trinomials cover most cases. Example in miniature: $\dfrac{x^2-16}{x-4}=\dfrac{(x-4)(x+4)}{x-4}=x+4$, whose limit at $4$ is $8$.

2 · Rationalise with a conjugate. When a square root causes the $\tfrac{0}{0}$, multiply top and bottom by the conjugate — the same two terms with the sign between them flipped. The product is a difference of squares, which clears the root and reveals the cancelling factor. Choosing the same sign instead of the conjugate just squares you back to $\tfrac{0}{0}$, so the sign matters.

3 · Combine a complex fraction. When the numerator is itself a difference of fractions, add them over a common denominator first. The $(x-c)$ factor then appears in the numerator and cancels against the one in the denominator, and substitution finishes the job. Watch the sign when you rewrite $c-x$ as $-(x-c)$ — a lost minus sign flips the whole answer.

In every case the strategy is identical: the $\tfrac{0}{0}$ is telling you a factor is hidden. Do the algebra that reveals and cancels it, then substitute. And always verify at the end by mentally plugging a value close to $c$ (say $c+0.01$) into the original expression — the answer you found should be right beside it.

A two-sided limit exists only when the left-hand and right-hand limits agree. As $x$ approaches $c$ from below we write $\lim_{x\to c^-} f(x)$, and from above $\lim_{x\to c^+} f(x)$. When both one-sided limits close in on the same number $L$, the two-sided limit is $L$. When they disagree — a jump between two heights, or $+\infty$ on one side and $-\infty$ on the other — the two-sided limit does not exist, even though each one-sided limit may be perfectly well defined. That mismatch is exactly what happened at the non-zero-over-zero asymptote above: the fraction shot to $+\infty$ on the right and $-\infty$ on the left, so no single number could be the two-sided limit.

Faced with any limit, run the same short routine every time. First, substitute $x=c$. If you get an ordinary number, that number is the limit — stop, and do not invent extra algebra. If you get $\tfrac{0}{0}$, a factor is hidden: factor-and-cancel for polynomials, the conjugate for roots, a common denominator for stacked fractions — then substitute again. If instead you get a non-zero number over zero, it is an infinite limit: check the sign of the fraction from each side and report a vertical asymptote, with the two-sided limit not existing. Finally, verify by evaluating the original expression at a value a hair away from $c$. Following this routine turns a scary-looking limit into a two-minute, mechanical decision.

It is worth slowing down on the difference between the value and the limit one final time, because it is the idea students most often blur. In Figure 2 the function $(x^2-9)/(x-3)$ has no value at $x=3$ — substituting genuinely gives $\tfrac{0}{0}$, which is not a number. Yet the limit is a perfectly definite $6$, because the graph is a straight line heading straight for the point $(3,6)$ from both sides, and one missing point cannot change where a curve is pointed. A limit is patient: it watches the approach and ignores the destination.

When the algebra is not obvious, you can always estimate a limit by building a short table of $f(x)$ for $x$ marching toward $c$ from each side — say $2.9, 2.99, 2.999$ and $3.1, 3.01, 3.001$. If both columns close in on the same number, that number is almost certainly the limit, and it gives you a target to check your algebra against. A numerical table never proves a limit — it cannot rule out something strange happening even closer in — but it is an excellent sanity check and a fast way to catch a sign error or a botched cancellation before it costs you marks.

The three algebraic moves clear a hidden factor, but some limits have no factor to clear — they wobble. For these the Squeeze Theorem rescues the answer. If $g(x)\le f(x)\le h(x)$ for all $x$ in a punctured neighbourhood of $c$, and the outer functions are trapped together so that $\lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L$, then the middle function has nowhere to go and $\lim_{x\to c}f(x)=L$ as well. The showcase case is $\lim_{x\to 0}x^2\sin\tfrac1x$: the sine term oscillates without ever settling and has no limit on its own, yet $-x^2\le x^2\sin\tfrac1x\le x^2$, and both envelopes head to $0$, so the product is pinned to $0$. The same idea, applied to a unit-circle picture, is the standard route to $\lim_{x\to 0}\tfrac{\sin x}{x}=1$ — the limit you will need before you can differentiate the trigonometric functions.

The laws extend to composition, with one continuity condition. If $\lim_{x\to c}g(x)=L$ and the outer function $f$ is continuous at $L$, then $\lim_{x\to c}f\big(g(x)\big)=f(L)$ — you may carry the limit inside. This is exactly why $\lim_{x\to 2}\sqrt{x^2+5}=\sqrt{\lim_{x\to 2}(x^2+5)}=\sqrt{9}=3$: the square root is continuous at $9$, so the limit slides through it. The continuity caveat is real — for an outer function with a jump at $L$ the move can fail — but the polynomial, rational and root outer functions of this course are continuous wherever they are defined, so once the inside limit exists the substitution is legitimate.

Functions stitched together from different formulas are where the two-sided rule earns its keep. A piecewise function has a two-sided limit at a boundary point $c$ only when its two pieces arrive at the same height from either side. Take $f(x)=x+1$ for $x<1$ and $f(x)=4-x$ for $x\ge 1$: the left piece approaches $2$ while the right piece approaches $3$, so $\lim_{x\to 1}f(x)$ does not exist even though each one-sided limit is perfectly well defined. The lesson is procedural — always evaluate both sides of a boundary point separately before you declare a two-sided limit — and it is the exact habit the next lesson on one-sided limits, and later the test for continuity, are built on.

It is worth stating precisely why cancelling a factor is not cheating. Two functions that agree at every point of an interval except possibly one have the same limit at that point, because a limit is fixed entirely by the punctured neighbourhood $0<|x-c|<\delta$ and never by the value at $c$ itself. Factor-and-cancel replaces $\dfrac{(x-c)\,q(x)}{(x-c)\,r(x)}$ with $\dfrac{q(x)}{r(x)}$, and those two expressions are equal everywhere except at $x=c$ — precisely the one point the limit ignores. That agreement-off-a-point theorem is the justification underneath the whole technique, and it is why the simplified expression delivers the true limit despite being, strictly, a different function.

A word on the assumptions behind these rules, because knowing their boundaries matters as much as knowing the rules themselves. The limit laws apply only when the individual limits they combine actually exist and are finite; if one piece has no limit, the law that uses it says nothing at all, and the expression must be analysed directly instead. The quotient law carries the explicit exception that the denominator limit must be non-zero, and the even-root law the exception that the radicand limit must be non-negative. Direct substitution is valid only when the function is continuous at the point, so it cannot be applied to a rational function at a zero of its denominator. And the factor-and-cancel technique works only when a genuine common factor is present; it never manufactures one where none exists, so a non-zero number over zero must be treated as an infinite limit rather than an indeterminate form.

It is worth being equally explicit about what these algebraic techniques do not do. They resolve the indeterminate form that arises from a shared vanishing factor, but they are powerless against other indeterminate forms, such as a difference of two quantities that both grow without bound, or a product of a vanishing factor with an exploding one; those require their own methods met later. A successful cancellation also guarantees nothing about the original function value at the point, since the simplified expression agrees with the original only away from the hole, and the function itself remains undefined there. These techniques compute the limit, a statement purely about the approach, and deliberately say nothing about the destination the function may or may not reach.

Finally, every rule here is stated for a real-valued function of a single real variable approaching a finite point, which is the whole setting of this lesson. The behaviour of a function as its input grows without bound is a separate question, handled by the limits-at-infinity techniques of a later lesson, and limits of functions of several variables belong to a later course entirely. Within its intended domain the toolkit is complete and dependable, but recognising that domain, a finite point, a single real variable, and existing finite component limits, is exactly what keeps you from misapplying a rule outside the conditions that make it true in the first place.

None of this is busywork. The central object of the entire course — the derivative — is defined as a limit of exactly this indeterminate shape: $f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$, which is $\tfrac{0}{0}$ at $h=0$ for every function. Every derivative you ever compute from first principles is a $\tfrac{0}{0}$ limit resolved by the same factoring, expanding and cancelling you are practising here. Master removing $\tfrac{0}{0}$ now, and the definition of the derivative becomes a familiar friend instead of a wall.