CommeUnJeu · Grade 12 Pure Maths

Limits

⌚ ~162 min ▢ 18 blocks ✓ 62 exercises

Master the foundational concept of limits. This module covers numerical and algebraic evaluation, limit laws, one-sided limits, infinite behavior (asymptotes), the Squeeze Theorem and composite function analysis.

I Definition

Definition — Limit

A function $f$ has the limit $L$ as $x$ approaches $a$ if the values of $f(x)$ become arbitrarily close to the single real number $L$ whenever $x$ is sufficiently close to $a$ (but not necessarily equal to $a$), from both sides. We write this as:$$ \textcolor{colordef}{\lim_{x \to a} f(x) = L}\quad\text{or}\quad\textcolor{colordef}{f(x) \xrightarrow[x \to a]{} L}\quad\text{or}\quad\textcolor{colordef}{\text{as } x \to a,\ f(x)\to L}$$

The crucial idea of a limit is that it describes the behaviour of a function near a point, not at the point itself. The value of $f(a)$, or whether it even exists, is irrelevant to the value of the limit.
Consider the function $f(x) = \dfrac{x^2-1}{x-1}$. At $x=1$, the function is undefined. However, for any $x \neq 1$, we can simplify:$$ f(x) = \dfrac{(x-1)(x+1)}{x-1} = x+1. $$The graph of $f(x)$ is the line $y=x+1$ with a "hole" at $x=1$. As $x$ gets very close to 1 from either side, the value of $f(x)$ gets very close to 2. Therefore, $\displaystyle\lim_{x \to 1} f(x) = 2$, even though $f(1)$ is not defined.

Skills to practice

Investigating Limits Numerically

II Existence of a Limit

Definition — One-Sided Limits

The right-hand limit, $\displaystyle\lim_{x \to a^+} f(x)$, is the value that $f(x)$ approaches as $x$ approaches $a$ from values greater than $a$.
The left-hand limit, $\displaystyle\lim_{x \to a^-} f(x)$, is the value that $f(x)$ approaches as $x$ approaches $a$ from values less than $a$.

Proposition — Existence of a Limit

The two-sided limit $\displaystyle\lim_{x \to a} f(x)$ exists if and only if the left-hand and right-hand limits both exist and are equal:$$ \displaystyle\lim_{x \to a} f(x) = L \iff \displaystyle\lim_{x \to a^-} f(x) = L \text{ and } \displaystyle\lim_{x \to a^+} f(x) = L $$

Example

Consider the piecewise function $f(x) = \begin{cases} x, & x < 1 \\ x^2, & x \ge 1 \end{cases}$. Does $\lim_{x \to 1} f(x)$ exist?

Answer

To determine if the limit exists, we must find the left-hand and right-hand limits and check if they are equal.

Left-hand limit: As $x \to 1^-$, we use the rule for $x<1$.$$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x) = 1 $$
Right-hand limit: As $x \to 1^+$, we use the rule for $x \ge 1$.$$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1 $$

Since the left-hand limit (1) is equal to the right-hand limit (1), the limit exists and $\lim_{x \to 1} f(x) = 1$.

Example

Consider the piecewise function $f(x) = \begin{cases} x+1, & x < 1 \\ x^2, & x \ge 1 \end{cases}$. Does $\lim_{x \to 1} f(x)$ exist?

Answer

To determine if the limit exists, we must find the left-hand and right-hand limits and check if they are equal.

Left-hand limit: As $x \to 1^-$, we use the rule for $x<1$.$$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+1) = 1+1 = 2 $$
Right-hand limit: As $x \to 1^+$, we use the rule for $x \ge 1$.$$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1 $$

Since the left-hand limit (2) is not equal to the right-hand limit (1), the limit $\lim_{x \to 1} f(x)$ does not exist. The graph shows a "jump" at $x=1$.

Skills to practice

Evaluating Limits Graphically

III Infinite Limits and Vertical Asymptotes

We have seen that a limit fails to exist if the function approaches different values from the left and right. Another way a limit can fail to exist is if the function's values grow without bound, approaching positive or negative infinity. While these limits do not technically exist as finite numbers, we use limit notation as a precise way to describe this "unbounded behaviour," which corresponds to a vertical asymptote on a graph.

Definition — Infinite Limit

The notation $\displaystyle\lim_{x \to a} f(x) = +\infty$ means that the values of $f(x)$ can be made arbitrarily large and positive by taking $x$ sufficiently close to $a$ (from both sides).
Similarly for $\displaystyle\lim_{x \to a} f(x) = -\infty$.
In this situation we say that the limit diverges to $+\infty$ or $-\infty$; it is not a finite real number.

Definition — Vertical Asymptote

The line $x=a$ is a vertical asymptote of the graph of the function $f(x)$ if the function approaches $+\infty$ or $-\infty$ as $x$ approaches $a$ from the left or the right.

Example

Investigate $\displaystyle\lim_{x \to 0} \dfrac{1}{x^2}$.

Answer

Direct substitution of $x=0$ results in $\dfrac{1}{0}$, which is undefined, suggesting a vertical asymptote at $x=0$. We check the one-sided limits:

As $x \to 0^+$, $x^2$ is a small positive number, so $\dfrac{1}{x^2} \to +\infty$.
As $x \to 0^-$, $x^2$ is also a small positive number, so $\dfrac{1}{x^2} \to +\infty$.

Since the function's values increase without bound towards $+\infty$ from both sides, we describe this behavior as $\displaystyle\lim_{x \to 0} \dfrac{1}{x^2} = +\infty$.

Skills to practice

Investigating Limits Numerically
Evaluating Infinite Limits Graphically

IV Limits at Infinity

Limits at infinity describe the end behaviour of a function. When $\displaystyle\lim_{x \to \infty} f(x) = L$ (or $\displaystyle\lim_{x \to -\infty} f(x) = L$), the line $y=L$ is a horizontal asymptote of the graph.

Definition — Limit at Infinity

A function $f$ has the limit $L$ as $x$ approaches $+\infty$ when the values of $f(x)$ become arbitrarily close to the single real number $L$ for all sufficiently large $x$. We write:$$ \textcolor{colordef}{\lim_{x \to +\infty} f(x) = L}\quad\text{or}\quad\textcolor{colordef}{f(x) \xrightarrow[x \to +\infty]{} L}. $$

A similar definition applies for $x \to -\infty$.

Definition — Horizontal Asymptote

The line $y=L$ is a horizontal asymptote of the graph of the function $f(x)$ if either of the following limit conditions is met:$$ \lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L $$

Skills to practice

Investigating Limits Numerically
Evaluating End Behavior Graphically

V Reference Limits and Operations

Proposition — Reference Limits

The following limits for standard functions are used as building blocks for more complex calculations:

Function	$\mathbf{x \to -\infty}$	$\mathbf{x \to +\infty}$	$\mathbf{x \to 0^-}$	$\mathbf{x \to 0^+}$
$1/x$	$0$	$0$	$-\infty$	$+\infty$
$x^n$ ($n \in \mathbb{N}^*$)	$+\infty$ (even) / $-\infty$ (odd)	$+\infty$	0	0
$e^x$	$0$	$+\infty$	1	1
$\sqrt{x}$	Undefined	$+\infty$	Undefined	$0$
$1/\sqrt{x}$	Undefined	$0$	Undefined	$+\infty$

While tables of values and graphs are useful for understanding the concept of a limit, they are not efficient for evaluation. Fortunately, limits follow a set of predictable algebraic rules, known as the Limit Laws, which allow us to calculate them directly.

Proposition — Operations on Limits

Let $L$ and $L'$ be real numbers. The following tables summarize the rules for limits of sums, products, and quotients involving infinity.

Sum and Product:

$\lim f(x)$	$\lim g(x)$	$\lim (f(x)+g(x))$	$\lim (f(x) \times g(x))$
$L$	$L'$	$L+L'$	$L \times L'$
$L\neq 0$	$\pm \infty$	$\pm \infty$	$\pm \infty$ (sign rule)
$\pm \infty$	$\pm \infty$	$\pm \infty$ (if same sign)	$\pm \infty$ (sign rule)
$0$	$\pm \infty$	$\pm \infty$	Indeterminate Form
$+\infty$	$-\infty$	Indeterminate Form	$-\infty$

Quotient:

$\lim f(x)$	$\lim g(x)$	$\lim \dfrac{f(x)}{g(x)}$
$L$	$L' \neq 0$	$\dfrac{L}{L'}$
$L$	$\pm \infty$	$0$
$L \neq 0$	$0$	$\pm \infty$ (sign rule)
$\pm \infty$	$L' \neq 0$	$\pm \infty$ (sign rule)
$0$	$0$	Indeterminate Form
$\pm \infty$	$\pm \infty$	Indeterminate Form

Method — Evaluating Limits

Direct Substitution (when applicable): Always start by substituting $x=a$ into $f(x)$.
- If the expression is defined at $x=a$ (no division by $0$, no square root of a negative number, etc.) and the function is continuous at $a$ (as is the case for polynomials, rational functions where the denominator is nonzero, $e^x$, $\ln x$ on its domain, etc.), then:$$\lim_{x\to a} f(x)=f(a).$$
- If the expression is not defined, or if the function may have a discontinuity, direct substitution is not sufficient: we must transform the expression or study one-sided limits.
Algebraic Manipulation: If substitution leads to an indeterminate form such as $\dfrac{0}{0}$, simplify the expression by:
- factoring and cancelling common factors (when allowed);
- using a conjugate (for expressions with roots);
- rewriting the expression (common denominator, factoring out dominant terms, etc.).
After simplifying, try direct substitution again.

Caution: Direct substitution is reliable for continuous functions. However, if a function has a discontinuity (a “jump”), a hole, or different formulas on either side of a point (such as the floor function $\lfloor x \rfloor$), substitution can be misleading. In that case, study one-sided limits or transform the expression first.

Skills to practice

Evaluating Limits by Direct Substitution
Applying the Limit Laws
Evaluating Limits using Operations
Justifying Limits using Operations
Justifying One-Sided Limits using Operations
Determining One-Sided Limits and Vertical Asymptotes
Resolving Indeterminate Forms
Evaluating Limits by Algebraic Simplification
Evaluating Limits at Infinity by Algebraic Simplification
Finding Derivatives from First Principles

VI Limit of a Composite Function

Proposition — Limit of a Composite Function

Let $a, L$ and $M$ be real numbers or $\pm\infty$.
If $\displaystyle\lim_{x \to a} g(x) = L$ and if $\displaystyle\lim_{X \to L} f(X) = M$, then:$$ \lim_{x \to a} f(g(x)) = M. $$

In practice, to find the limit of $f(g(x))$, we first determine what the "inner" part $g(x)$ approaches, and then find the limit of $f$ at that value.

Example

Evaluate $\displaystyle\lim_{x \to +\infty} e^{\frac{1}{x}}$.

Answer

We decompose the function into an inner part $g(x)$ and an outer part $f(X)$:

Inner limit: $\displaystyle\lim_{x \to +\infty} \dfrac{1}{x} = 0$.
Outer limit: $\displaystyle\lim_{X \to 0} e^X = e^0 = 1$.

By the rule of composition, $\displaystyle\lim_{x \to +\infty} e^{\frac{1}{x}} = 1$.

Skills to practice

Evaluating Limits of Composite Functions

VII The Squeeze Theorem

Some limits, particularly those involving oscillating functions like $\sin(1/x)$, cannot be evaluated using direct substitution or standard algebraic manipulation alone. For these cases, we may be able to use the Squeeze Theorem (also known as the Sandwich Theorem). The idea is to find two simpler functions that "squeeze" or "sandwich" the more complex function between them. If these two outer functions approach the same limit, then the function trapped in the middle must also approach that same limit.

Proposition — The Squeeze Theorem

Suppose that for all $x$ in some interval around $a$ (except possibly at $x=a$):$$ g(x) \le f(x) \le h(x). $$And suppose that:$$ \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L. $$Then, we can conclude that:$$ \lim_{x \to a} f(x) = L. $$

Example

Evaluate $\displaystyle\lim_{x \to 0} x^2 \sin\left(\dfrac{1}{x}\right)$.

Answer

We cannot use direct substitution because $\sin(1/0)$ is undefined. The function $f(x) = x^2 \sin(1/x)$ is made of two parts: $x^2$, which approaches $0$, and $\sin(1/x)$, which oscillates infinitely between $-1$ and $1$ as $x \to 0$. This suggests using the Squeeze Theorem.
We start with the known range of the sine function:$$ -1 \le \sin\left(\dfrac{1}{x}\right) \le 1. $$Since $x^2 \ge 0$, we can multiply the entire inequality by $x^2$ without changing the direction of the inequality signs:$$ -x^2 \le x^2 \sin\left(\dfrac{1}{x}\right) \le x^2. $$We have now "squeezed" our function between $g(x)=-x^2$ and $h(x)=x^2$. Next, we find the limits of these outer functions as $x \to 0$:$$ \lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} (x^2) = 0. $$Since our function is trapped between two functions that both approach the limit $0$, by the Squeeze Theorem, our limit must also be $0$:$$ \lim_{x \to 0} x^2 \sin\left(\dfrac{1}{x}\right) = 0. $$

Skills to practice

Applying the Squeeze Theorem

VIII Review \& Beyond

Skills to practice

Quiz

Function	\(\mathbf{x \to -\infty}\)	\(\mathbf{x \to +\infty}\)	\(\mathbf{x \to 0^-}\)	\(\mathbf{x \to 0^+}\)
\(1/x\)	\(0\)	\(0\)	\(-\infty\)	\(+\infty\)
\(x^n\) (\(n \in \mathbb{N}^*\))	\(+\infty\) (even) / \(-\infty\) (odd)	\(+\infty\)	0	0
\(e^x\)	\(0\)	\(+\infty\)	1	1
\(\sqrt{x}\)	Undefined	\(+\infty\)	Undefined	\(0\)
\(1/\sqrt{x}\)	Undefined	\(0\)	Undefined	\(+\infty\)

\(\lim f(x)\)	\(\lim g(x)\)	\(\lim (f(x)+g(x))\)	\(\lim (f(x) \times g(x))\)
\(L\)	\(L'\)	\(L+L'\)	\(L \times L'\)
\(L\neq 0\)	\(\pm \infty\)	\(\pm \infty\)	\(\pm \infty\) (sign rule)
\(\pm \infty\)	\(\pm \infty\)	\(\pm \infty\) (if same sign)	\(\pm \infty\) (sign rule)
\(0\)	\(\pm \infty\)	\(\pm \infty\)	Indeterminate Form
\(+\infty\)	\(-\infty\)	Indeterminate Form	\(-\infty\)

\(\lim f(x)\)	\(\lim g(x)\)	\(\lim \dfrac{f(x)}{g(x)}\)
\(L\)	\(L' \neq 0\)	\(\dfrac{L}{L'}\)
\(L\)	\(\pm \infty\)	\(0\)
\(L \neq 0\)	\(0\)	\(\pm \infty\) (sign rule)
\(\pm \infty\)	\(L' \neq 0\)	\(\pm \infty\) (sign rule)
\(0\)	\(0\)	Indeterminate Form
\(\pm \infty\)	\(\pm \infty\)	Indeterminate Form