Limits

What the notation means. Writing $\lim_{x\to a} f(x) = L$ says that as $x$ takes values closer and closer to $a$ — but never equal to $a$ — the outputs $f(x)$ crowd closer and closer to one number $L$. The limit is a statement about the neighbourhood of $a$, never about the single point $a$.

Two-sided agreement. A limit exists at $a$ only when the left-hand limit (approaching from below) and the right-hand limit (approaching from above) both exist and match:

where: $x \to a^-$ means $x$ approaches $a$ through values less than $a$, and $x \to a^+$ through values greater than $a$. If the two sides disagree, the two-sided limit simply does not exist — there is no "average" fallback.

Algebra of limits. If $\lim_{x\to a} f(x) = l$ and $\lim_{x\to a} g(x) = m$ both exist, limits combine the way you would hope:

where: $l$ and $m$ are the individual limits and the quotient rule requires the denominator's limit to be nonzero. For a polynomial $p(x)$, this collapses to plain substitution: $\lim_{x\to a} p(x) = p(a)$.

When substitution fails. Direct substitution into a rational function $\frac{p(x)}{q(x)}$ can produce the indeterminate form $\frac{0}{0}$ when $q(a) = 0$. This is not a dead end — it is a signal that $p(x)$ and $q(x)$ share a common factor of $(x-a)$ that must be cancelled before the limit can be read off. Note also that $\frac{0}{0}$ is very different from a genuinely nonexistent limit: if substitution instead gives a nonzero number over zero, such as $\frac{5}{0}$, the function is blowing up near that point and no amount of algebra will produce a finite limit — that case is not covered by this chapter's techniques. The standard shortcut for one common family of $\frac00$ limits:

where: $n$ is a positive integer power and $a$ is the point being approached. Three further standard results, quoted without proof at this syllabus level, resolve limits involving trigonometric, exponential and logarithmic functions:

Evaluate $\displaystyle\lim_{x \to 0} \frac{\sqrt{4+x}-2}{x}$.

Step 1. Try direct substitution first. At $x=0$: numerator $= \sqrt4 - 2 = 0$, denominator $= 0$. This is the indeterminate form $\frac00$, and there is no common polynomial factor visible yet because of the square root — so the fix here is rationalising, not plain factoring.

Step 2. Multiply numerator and denominator by the conjugate $\sqrt{4+x}+2$:

Step 3. Cancel the common factor of $x$, valid because $x \to 0$ means $x \neq 0$ throughout the limiting process:

Step 4. Now substitution poses no difficulty since the square root in the denominator no longer multiplies against a vanishing numerator:

So $\displaystyle\lim_{x\to 0}\frac{\sqrt{4+x}-2}{x} = \dfrac14$.

Competitive stretch

Check whether $\displaystyle\lim_{x\to 1} f(x)$ exists for $f(x) = 3x+1$ when $x<1$ and $f(x) = 5x-1$ when $x \geq 1$. The left-hand limit is $3(1)+1 = 4$ and the right-hand limit is $5(1)-1 = 4$; the two sides agree, so $\lim_{x\to 1} f(x) = 4$, exactly as in the standard piecewise check — a reminder that the limit only depends on values near $x=1$, never on the defining rule at $x=1$ itself.

• Assuming $\lim_{x\to a} f(x)$ always equals $f(a)$ — true for polynomials at ordinary points, but the limit can exist even when $f(a)$ is undefined, as the graph above shows.
• Seeing $\frac00$ and declaring the limit "does not exist" — $\frac00$ is indeterminate, a signal to factor or rationalise further, not a dead end.
• Skipping the one-sided check on piecewise functions; the two-sided limit exists only when left-hand and right-hand limits agree.
• Cancelling $(x-a)$ without noting that this is valid precisely because $x \neq a$ throughout the limiting process, never at $x=a$ itself.

Ask the tutor: If a $\frac00$ limit will not simplify no matter which technique you try, or you are unsure whether a piecewise function's one-sided limits actually agree, ask the AI tutor to work through it with you step by step.