Limits and Derivatives • Topic 1 of 3

Intuitive Idea and Algebra of Limits

A limit describes the value a function approaches as its input creeps towards some number $a$ — not the value at $a$, which the function may never actually reach (or may not even be defined there). We write $\lim_{x \to a} f(x) = L$ to mean: as $x$ gets arbitrarily close to $a$, the output $f(x)$ gets arbitrarily close to $L$.

One-sided limits. Because $x$ can approach $a$ from two directions, we distinguish the left-hand limit (LHL), approaching through values smaller than $a$, and the right-hand limit (RHL), approaching through larger values:

$$\text{LHL} = \lim_{x \to a^-} f(x), \qquad \text{RHL} = \lim_{x \to a^+} f(x)$$

Existence of a limit. The limit at $a$ exists if and only if both one-sided limits exist and agree. This is the single most-tested idea of the topic:

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

If the two sides disagree — as happens at a jump — the limit simply does not exist, no matter how nicely the function behaves elsewhere.

Algebra of limits. Provided $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist, limits respect the basic operations. Let $\lim f = L$ and $\lim g = M$:

LawStatement
Sum / Difference$\lim (f \pm g) = L \pm M$
Constant multiple$\lim (k\,f) = kL$
Product$\lim (f \cdot g) = L \cdot M$
Quotient$\lim \dfrac{f}{g} = \dfrac{L}{M}, \;\; M \ne 0$
Power$\lim \big(f\big)^{n} = L^{n}$

Polynomials and rational functions. Polynomials are continuous everywhere, so for any polynomial $p(x)$ you get the limit by direct substitution: $\lim_{x \to a} p(x) = p(a)$. For a rational function $\dfrac{p(x)}{q(x)}$, substitution works as long as $q(a) \ne 0$.

The $\dfrac{0}{0}$ form. When both numerator and denominator vanish at $a$, substitution gives the meaningless symbol $\dfrac{0}{0}$ — an indeterminate form. The cure is to factorise and cancel the common factor $(x - a)$ that is causing both to vanish, then substitute into what remains:

$$\lim_{x \to 2}\dfrac{x^2 - 4}{x - 2} = \lim_{x \to 2}\dfrac{(x-2)(x+2)}{x-2} = \lim_{x \to 2}(x+2) = 4$$

Deeper Insight — a limit is about the journey, not the destination: The whole power of the limit concept comes from a deliberate separation between what a function does near a point and what it is at that point. In the example above, $\dfrac{x^2-4}{x-2}$ is genuinely undefined at $x = 2$ — there is a hole in the graph — yet the limit is a perfectly definite $4$, because every nearby value of the function is close to $4$. This is exactly why cancelling the factor $(x-2)$ is legitimate: for all $x \ne 2$ the two expressions are identical, and the limit only ever cares about $x \ne a$. The $\dfrac{0}{0}$ symbol is not an answer but a signal that a hidden common factor is masking the true behaviour, and learning to read that signal — factor, rationalise, or simplify — is the skill that the entire calculus syllabus is built upon.

Function approaching a limit value L from both sides A Limit: f(x) → L as x → a La from left →← from right LHL = RHL = L, even though f(a) may be undefined Removable hole created by a zero over zero form 0/0 form: a removable hole 42 y = (x²−4)/(x−2) undefined at x = 2, but limit = 4
1
Worked Example
Evaluate $\lim_{x \to 3}\,(2x^2 - 5x + 1)$.
Solution
  1. A polynomial is continuous everywhere, so use direct substitution.
  2. $2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4$.

Answer: $\lim_{x \to 3}\,(2x^2 - 5x + 1) = 4$.

2
Worked Example
Evaluate $\lim_{x \to 1}\dfrac{x^2 - 1}{x - 1}$.
Solution
  1. Substituting $x = 1$ gives $\dfrac{0}{0}$ — an indeterminate form.
  2. Factorise the numerator: $x^2 - 1 = (x - 1)(x + 1)$.
  3. Cancel the common factor $(x - 1)$: $\dfrac{(x-1)(x+1)}{x-1} = x + 1$ for $x \ne 1$.
  4. Now substitute: $\lim_{x \to 1}(x + 1) = 2$.

Answer: $\lim_{x \to 1}\dfrac{x^2 - 1}{x - 1} = 2$.

3
Worked Example
Test whether $\lim_{x \to 0} f(x)$ exists for $f(x) = \begin{cases} x + 2, & x < 0 \\ 3 - x, & x \ge 0 \end{cases}$.
Solution
  1. Left-hand limit: $\lim_{x \to 0^-}(x + 2) = 0 + 2 = 2$.
  2. Right-hand limit: $\lim_{x \to 0^+}(3 - x) = 3 - 0 = 3$.
  3. Since LHL $= 2 \ne 3 = $ RHL, the one-sided limits disagree.

Answer: The limit does not exist at $x = 0$ (a jump).

4
Worked Example
Evaluate $\lim_{x \to 2}\dfrac{x^2 - 5x + 6}{x^2 - 4}$.
Solution
  1. Substituting $x = 2$ gives $\dfrac{0}{0}$, so factorise both parts.
  2. Numerator: $x^2 - 5x + 6 = (x - 2)(x - 3)$.
  3. Denominator: $x^2 - 4 = (x - 2)(x + 2)$.
  4. Cancel $(x - 2)$: $\dfrac{x - 3}{x + 2}$, then substitute: $\dfrac{2 - 3}{2 + 2} = -\dfrac{1}{4}$.

Answer: $\lim_{x \to 2}\dfrac{x^2 - 5x + 6}{x^2 - 4} = -\dfrac{1}{4}$.

5
Worked Example
Evaluate $\lim_{x \to 0}\dfrac{\sqrt{1 + x} - 1}{x}$.
Solution
  1. Direct substitution gives $\dfrac{0}{0}$; rationalise by multiplying by the conjugate $\sqrt{1+x}+1$.
  2. $\dfrac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} = \dfrac{(1+x) - 1}{x(\sqrt{1+x}+1)} = \dfrac{x}{x(\sqrt{1+x}+1)}$.
  3. Cancel $x$: $\dfrac{1}{\sqrt{1+x}+1}$, then substitute $x = 0$: $\dfrac{1}{1 + 1} = \dfrac{1}{2}$.

Answer: $\lim_{x \to 0}\dfrac{\sqrt{1 + x} - 1}{x} = \dfrac{1}{2}$.

6
Worked Example
Given $\lim_{x \to a} f(x) = 4$ and $\lim_{x \to a} g(x) = -2$, find $\lim_{x \to a}\dfrac{3f(x) + g(x)}{f(x)\,g(x)}$.
Solution
  1. Apply the algebra of limits term by term.
  2. Numerator: $3(4) + (-2) = 12 - 2 = 10$.
  3. Denominator: $f \cdot g \to (4)(-2) = -8$, which is non-zero, so the quotient law applies.
  4. Divide: $\dfrac{10}{-8} = -\dfrac{5}{4}$.

Answer: $\lim_{x \to a}\dfrac{3f(x) + g(x)}{f(x)\,g(x)} = -\dfrac{5}{4}$.

Key Points

  • A limit captures what $f(x)$ approaches near $a$, not its value at $a$.
  • $\lim_{x \to a} f(x)$ exists only if LHL $=$ RHL.
  • For polynomials and well-defined rational functions, use direct substitution.
  • The $\dfrac{0}{0}$ form is indeterminate — factorise and cancel (or rationalise) the common factor.
  • Limits distribute over sums, products and quotients provided each piece exists (denominator $\ne 0$).
Tap an option to check your answer0 / 4
Q1.$\displaystyle\lim_{x\to2}(x+3)=$
Explanation: Substitute $x=2$.
Q2.$\displaystyle\lim_{x\to a} c=$ (where $c$ is constant):
Explanation: Limit of a constant is the constant.
Q3.$\displaystyle\lim_{x\to0} x^2=$
Explanation: Substitute $x=0$.
Q4.$\displaystyle\lim_{x\to a}\big(f(x)+g(x)\big)=$
Explanation: Sum law of limits.