Limits and Continuity | AP - Wyatt's Notes

Intuitive Notion of a Limit (CED Unit 1)

The limit of a function $f(x)$ as $x$ approaches $a$ is the value that $f(x)$ approaches, regardless Of whether $f(a)$ is defined:

\lim_{x \to a} f(x) = L

This means that as $x$ gets arbitrarily close to $a$ , $f(x)$ gets arbitrarily close to $L$ .

Left-Hand and Right-Hand Limits

A two-sided limit exists if and only if both one-sided limits exist and are equal:

\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L

Example Find $\displaystyle\lim_{x \to 3} \frac{|x - 3|}{x - 3}$ .

For $x \lt 3$ : $\frac{|x-3|}{x-3} = \frac{3-x}{x-3} = -1$ .

For $x \gt 3$ : $\frac{|x-3|}{x-3} = \frac{x-3}{x-3} = 1$ .

Left limit $= -1$ Right limit $= 1$ . The limit does not exist.

Common Limits

Limit	Value
$\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$	$1$
$\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x}$	$0$
$\displaystyle\lim_{x \to \infty} \frac{1}{x}$	$0$
$\displaystyle\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$	$e$
$\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}$	$1$
$\displaystyle\lim_{x \to 0} \frac{\ln(1+x)}{x}$	$1$

Proof that $\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

This is proved using the squeeze theorem and the geometric inequality $\sin x \lt x \lt \tan x$ for $0 \lt x \lt \frac{\pi}{2}$ Which gives $\cos x \lt \frac{\sin x}{x} \lt 1$ .

As $x \to 0^+$ , $\cos x \to 1$ So by the squeeze theorem, $\frac{\sin x}{x} \to 1$ . A similar Argument applies from the left. $\blacksquare$

Proof of the geometric inequality $\sin x \lt x \lt \tan x$ for $0 \lt x \lt \frac{\pi}{2}$ . Consider a unit circle sector with angle $x$ . The area of triangle $OAP$ (with altitude $\sin x$ ) is $\frac{1}{2}\sin x$ The area of the sector is $\frac{1}{2}x$ And the area of triangle $OAT$ (with Altitude $\tan x$ ) is $\frac{1}{2}\tan x$ . Since the sector contains the first triangle and is Contained in the second, we get $\frac{1}{2}\sin x \lt \frac{1}{2}x \lt \frac{1}{2}\tan x$ Hence $\sin x \lt x \lt \tan x$ . $\blacksquare$

Proof that $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ .

Let $h = e^x - 1$ So $e^x = 1 + h$ and $x = \ln(1+h)$ . As $x \to 0$ , $h \to 0$ .

\frac{e^x - 1}{x} = \frac{h}{\ln(1+h)} = \frac{1}{\frac{\ln(1+h)}{h}}

Since $\displaystyle\lim_{h \to 0} \frac{\ln(1+h)}{h} = 1$ (which follows from $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ and the fact that $\ln$ and $\exp$ are Inverses), we obtain the result. $\blacksquare$

Proof that $\displaystyle\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$ .

Let $u = \ln(1+x)$ So $e^u = 1 + x$ and $x = e^u - 1$ . As $x \to 0$ , $u \to 0$ .

\frac{\ln(1+x)}{x} = \frac{u}{e^u - 1} = \frac{1}{\frac{e^u - 1}{u}}

Since $\displaystyle\lim_{u \to 0} \frac{e^u - 1}{u} = 1$ The result follows. $\blacksquare$

The Squeeze Theorem

If $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$ ), and:

\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L

Then $\displaystyle\lim_{x \to a} f(x) = L$ .

Intuition: If $f$ is sandwiched between two functions that both approach $L$ Then $f$ must also Approach $L$ . The squeeze theorem is particularly useful when $f$ oscillates or is otherwise hard to Evaluate directly.

Example Show that $\displaystyle\lim_{x \to 0} x^2 e^{\sin(1/x)} = 0$ .

Since $-1 \le \sin(1/x) \le 1$ We have $e^{-1} \le e^{\sin(1/x)} \le e^1$ .

Therefore:

E^{-1} x^2 \le x^2 e^{\sin(1/x)} \le e \cdot x^2

Both $\displaystyle\lim_{x \to 0} e^{-1} x^2 = 0$ and $\displaystyle\lim_{x \to 0} e \cdot x^2 = 0$ .

By the squeeze theorem, the limit is $0$ .

Algebraic Limit Properties

If $\displaystyle\lim_{x \to a} f(x) = L$ and $\displaystyle\lim_{x \to a} g(x) = M$ Then:

$\displaystyle\lim_{x \to a} [f(x) + g(x)] = L + M$
$\displaystyle\lim_{x \to a} [f(x) - g(x)] = L - M$
$\displaystyle\lim_{x \to a} [c \cdot f(x)] = cL$
$\displaystyle\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
$\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ Provided $M \ne 0$

Theorem (Limit of a power). If $\displaystyle\lim_{x \to a} f(x) = L$ and $n$ is a positive Integer, then $\displaystyle\lim_{x \to a} [f(x)]^n = L^n$ .

Theorem (Limit of a root). If $\displaystyle\lim_{x \to a} f(x) = L$ and $n$ is a positive Integer, and $L \ge 0$ when $n$ is even, then $\displaystyle\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$ .

Proof of property 1 (sum rule). We need to show that for every $\epsilon \gt 0$ There exists $\delta \gt 0$ such that $|x - a| \lt \delta$ implies $|(f+g)(x) - (L+M)| \lt \epsilon$ .

Note that $|(f+g)(x) - (L+M)| = |(f(x) - L) + (g(x) - M)| \le |f(x) - L| + |g(x) - M|$ .

Evaluating Limits

Direct Substitution

When a function is continuous at $a$ The limit equals the function value:

\lim_{x \to a} f(x) = f(a)

Limits Involving Infinity

For rational functions $\displaystyle\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials:

If $\deg P \lt \deg Q$ : $\displaystyle\lim_{x \to \pm\infty} \frac{P(x)}{Q(x)} = 0$
If $\deg P = \deg Q$ : $\displaystyle\lim_{x \to \pm\infty} \frac{P(x)}{Q(x)} = \frac{\mathrm{leading coeff of P}{\mathrm{leading coeff of Q}$
If $\deg P \gt \deg Q$ : the limit is $\pm\infty$

Why this works. For large $x$ The leading term dominates. Dividing numerator and denominator by The highest power of $x$ in the denominator, all lower-order terms vanish.

Example Find $\displaystyle\lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2}$ .

Combine the numerator:

\frac{\frac{1}{x} - \frac{1}{2}}{x - 2} = \frac{\frac{2 - x}{2x}}{x - 2} = \frac{-(x-2)}{2x(x-2)} = -\frac{1}{2x}

Therefore:

\lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} = -\frac{1}{4}

Formal Definition of a Limit (Epsilon-Delta)

The precise definition: $\displaystyle\lim_{x \to a} f(x) = L$ means that for every $\epsilon \gt 0$ There exists a $\delta \gt 0$ such that:

0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \epsilon

Intuition. Think of it as a game. Your opponent picks $\epsilon$ (how close $f(x)$ must be to $L$ ). You must respond with $\delta$ (how close $x$ must be to $a$ ). If you can always win this Game, the limit exists.

Proving a Limit with Epsilon-Delta

Example Prove that $\displaystyle\lim_{x \to 3} (2x - 1) = 5$ .

We need to show that for every $\epsilon \gt 0$ There exists $\delta \gt 0$ such that $0 \lt |x - 3| \lt \delta \implies |(2x-1) - 5| \lt \epsilon$ .

Working backwards: $|(2x-1) - 5| = |2x - 6| = 2|x - 3|$ .

We want $2|x - 3| \lt \epsilon$ So choose $\delta = \frac{\epsilon}{2}$ .

Proof. Let $\epsilon \gt 0$ . Choose $\delta = \frac{\epsilon}{2}$ . Then:

0 \lt |x - 3| \lt \delta \implies |x - 3| \lt \frac{\epsilon}{2} \implies 2|x - 3| \lt \epsilon \implies |(2x - 1) - 5| \lt \epsilon

Therefore, $\displaystyle\lim_{x \to 3} (2x - 1) = 5$ . $\blacksquare$

Example Prove that $\displaystyle\lim_{x \to 2} x^2 = 4$ .

We need $|x^2 - 4| \lt \epsilon$ whenever $0 \lt |x - 2| \lt \delta$ .

Note that $|x^2 - 4| = |x - 2| \cdot |x + 2|$ .

If we restrict $\delta \le 1$ Then $|x - 2| \lt 1$ So $1 \lt x \lt 3$ and $|x + 2| \lt 5$ .

Thus $|x^2 - 4| = |x - 2| \cdot |x + 2| \lt 5|x - 2|$ .

Choose $\delta = \min\!\left(1, \frac{\epsilon}{5}\right)$ .

Proof. Let $\epsilon \gt 0$ . Choose $\delta = \min\!\left(1, \frac{\epsilon}{5}\right)$ . If $0 \lt |x - 2| \lt \delta$ Then:

|x^2 - 4| = |x - 2| \cdot |x + 2| \lt \delta \cdot 5 \le \frac{\epsilon}{5} \cdot 5 = \epsilon

Therefore, $\displaystyle\lim_{x \to 2} x^2 = 4$ . $\blacksquare$

Example Prove that $\displaystyle\lim_{x \to a} \sqrt{x} = \sqrt{a}$ for $a \gt 0$ .

We need $|\sqrt{x} - \sqrt{a}| \lt \epsilon$ whenever $0 \lt |x - a| \lt \delta$ .

Rationalise: $|\sqrt{x} - \sqrt{a}| = \frac{|x - a|}{\sqrt{x} + \sqrt{a}} \le \frac{|x - a|}{\sqrt{a}}$ .

We want $\frac{|x - a|}{\sqrt{a}} \lt \epsilon$ So $|x - a| \lt \epsilon\sqrt{a}$ .

Choose $\delta = \min(a, \epsilon\sqrt{a})$ . The condition $\delta \le a$ ensures $x \gt 0$ so that $\sqrt{x}$ is defined.

Proof. Let $\epsilon \gt 0$ . Choose $\delta = \min(a, \epsilon\sqrt{a})$ . If $0 \lt |x - a| \lt \delta$ Then $x \gt 0$ and:

|\sqrt{x} - \sqrt{a}| = \frac{|x - a|}{\sqrt{x} + \sqrt{a}} \le \frac{|x - a|}{\sqrt{a}} \lt \frac{\epsilon\sqrt{a}}{\sqrt{a}} = \epsilon

Therefore, $\displaystyle\lim_{x \to a} \sqrt{x} = \sqrt{a}$ . $\blacksquare$

Strategy for Epsilon-Delta Proofs

The general approach is:

Start with $|f(x) - L|$ and try to bound it in terms of $|x - a|$ .
If $f$ involves products, use the “restrict delta” technique: bound each factor separately.
If $f$ involves roots, rationalise and use the fact that $\sqrt{x} + \sqrt{a} \ge \sqrt{a}$ .
Choose $\delta = \min(\mathrm{bound, \epsilon / \mathrm{constant)$ to handle both the restriction and the $\epsilon$ requirement.

Continuity (CED Unit 1)

A function $f$ is continuous at $a$ if all three conditions hold:

$f(a)$ is defined
$\displaystyle\lim_{x \to a} f(x)$ exists
$\displaystyle\lim_{x \to a} f(x) = f(a)$

Theorem. Every polynomial function is continuous everywhere. Every rational function is Continuous on its domain.

Theorem (Continuity of compositions). If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ Then $f \circ g$ is continuous at $a$ .

This theorem justifies statements like ” $\sqrt{x^2 + 1}$ is continuous everywhere” — $x^2 + 1$ is a Polynomial (continuous everywhere) and $\sqrt{x}$ is continuous at all positive values (and $x^2 + 1 \ge 1 \gt 0$ ).

Types of Discontinuities

Type	Description	Example
Removable	Limit exists but $f(a)$ is undefined or $f(a) \ne \lim_{x \to a} f(x)$	$f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$
Jump (Non-removable)	One-sided limits exist but are not equal	$f(x) = \lfloor x \rfloor$
Infinite (Non-removable)	Function approaches $\pm\infty$	$f(x) = \frac{1}{x}$ at $x = 0$
Oscillating	Function oscillates without approaching a single value	$f(x) = \sin\!\left(\frac{1}{x}\right)$ at $x=0$

Intermediate Value Theorem (IVT)

If $f$ is continuous on $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$ Then there exists At least one $c \in (a, b)$ such that $f(c) = k$ .

Extreme Value Theorem

If $f$ is continuous on a closed interval $[a, b]$ Then $f$ attains both an absolute maximum and an Absolute minimum on $[a, b]$ .

Boundedness Theorem

If $f$ is continuous on a closed interval $[a, b]$ Then $f$ is bounded on $[a, b]$ — that is, There exist real numbers $m$ and $M$ such that $m \le f(x) \le M$ for all $x \in [a, b]$ .

This follows directly from the EVT: the absolute minimum and maximum serve as the bounds.

Asymptotes and Limits at Infinity

Vertical Asymptotes

If $\displaystyle\lim_{x \to a^+} f(x) = \pm\infty$ or $\displaystyle\lim_{x \to a^-} f(x) = \pm\infty$ Then $x = a$ is a vertical asymptote.

For rational functions $\frac{P(x)}{Q(x)}$ Vertical asymptotes occur at zeros of $Q(x)$ that are Not also zeros of $P(x)$ (after cancellation).

Horizontal Asymptotes

If $\displaystyle\lim_{x \to \pm\infty} f(x) = L$ Then $y = L$ is a horizontal asymptote.
A function can have at most two horizontal asymptotes (one as $x \to \infty$ One as $x \to -\infty$ ).

Oblique (Slant) Asymptotes

If $\deg P = \deg Q + 1$ in a rational function, perform polynomial long division. The quotient (excluding remainder) gives the slant asymptote.

Example Find the horizontal asymptotes of $\displaystyle f(x) = \frac{3e^x}{e^x + 1}$ .

As $x \to \infty$ : Divide numerator and denominator by $e^x$ :

\frac{3}{1 + e^{-x}} \to \frac{3}{1 + 0} = 3

As $x \to -\infty$ : Divide numerator and denominator by $e^x$ :

\frac{3e^x}{e^x + 1} \to \frac{0}{0 + 1} = 0

Horizontal asymptotes: $y = 3$ (as $x \to \infty$ ) and $y = 0$ (as $x \to -\infty$ ).

L”Hopital’s Rule (CED BC and AB Unit 1.15)

If $\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)}$ produces the indeterminate form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ And $f$ and $g$ are differentiable near $a$ with $g'(x) \ne 0$ near $a$ Then:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

Provided the limit on the right exists.

When to use L’Hopital’s Rule. It applies ONLY to $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ Forms. Using it on a determinate form (e.g., $\frac{3}{5}$ ) is an error.

When L’Hopital’s Rule fails. If the limit $\displaystyle\lim_{x \to a} \frac{f'(x)}{g'(x)}$ does Not exist, this does NOT mean the original limit does not exist. L’Hopital’s Rule only gives a Conclusion when the right-hand limit exists (or is $\pm\infty$ ).

Example Let $f(x) = \begin{cases} x^2 + 1 & x \lt 2 \\ 3x - 1 & x \ge 2 \end{cases}$ .

Find $\displaystyle\lim_{x \to 2} f(x)$ and determine if $f$ is continuous at $x = 2$ .

Left-hand limit: $\displaystyle\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 + 1) = 5$ .

Right-hand limit: $\displaystyle\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 1) = 5$ .

Since both one-sided limits equal 5: $\displaystyle\lim_{x \to 2} f(x) = 5$ .

Check continuity: $f(2) = 3(2) - 1 = 5 = \lim_{x \to 2} f(x)$ .

Therefore, $f$ is continuous at $x = 2$ .

Common Pitfalls

Confusing the value of a function at a point with its limit. The limit at $a$ does not depend on $f(a)$ at all. A function can have a limit at a point where it is undefined.
Assuming $\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$ when the denominator limit is zero. This is invalid when the denominator limit is zero.
Forgetting to check both one-sided limits for piecewise functions and absolute values.
Misapplying L’Hopital’s Rule when the limit is not in indeterminate form. Always verify $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ before applying.
Claiming a limit exists when only one-sided limits are checked. Both must agree.
Using thousands separators in math mode. Write $1000000$ in math expressions, not $1,000,000$ .
Using angle brackets in math mode. Use $\lt$ and $\gt$ commands instead of < and >.
Forgetting the EVT requires a closed interval. Open intervals do not guarantee maxima/minima.
Assuming L’Hopital’s Rule always works. If $\displaystyle\lim \frac{f'(x)}{g'(x)}$ does not exist, you cannot conclude anything about the original limit. Try algebraic methods instead.
Applying the product rule for limits to indeterminate products. The limit $\displaystyle\lim_{x \to 0^+} x \ln x$ is not $0 \cdot (-\infty) = 0$ ; it requires rewriting as a quotient and applying L’Hopital’s Rule.
Forgetting the “restrict delta” step in epsilon-delta proofs for nonlinear functions. You must bound $|x - a|$ before bounding the other factors.

Practice Questions

Find $\displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$ by factoring.
Prove using the epsilon-delta definition that $\displaystyle\lim_{x \to 4} \sqrt{x} = 2$ .
Determine all points of discontinuity for $f(x) = \frac{x^2 + x - 6}{x^2 - 9}$ and classify each.
Find the horizontal and vertical asymptotes of $\displaystyle f(x) = \frac{3x^2 - 2x + 1}{x^2 - 4}$ .
Use L’Hopital’s Rule to find $\displaystyle\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$ .
Let $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & x \ne 3 \\ k & x = 3 \end{cases}$ . Find the value of $k$ that makes $f$ continuous at $x = 3$ .
Use the squeeze theorem to find $\displaystyle\lim_{x \to 0} x \cos\!\left(\frac{1}{x}\right)$ .
Given $f(x) = x^3 - 3x + 1$ Use the IVT to show there is at least one root in the interval $(1, 2)$ .
Find $\displaystyle\lim_{x \to 0} \frac{\tan x}{x}$ .
Find $\displaystyle\lim_{x \to 1} \frac{\sqrt{x} - 1}{\sqrt[3]{x} - 1}$ .
Classify each discontinuity of $\displaystyle f(x) = \frac{x^2 - x}{x^2 - 1}$ .
Use the IVT to prove that $f(x) = e^x - 3 - x$ has at least one root in the interval $(1, 2)$ .
Find $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$ .
Prove that $\displaystyle\lim_{x \to 3} \frac{1}{x} = \frac{1}{3}$ using the epsilon-delta definition.
Find the value of $c$ such that $f(x) = \begin{cases} cx^2 + 2x & x \lt 1 \\ 3x - 1 & x \ge 1 \end{cases}$ is continuous at $x = 1$ .
Evaluate $\displaystyle\lim_{x \to 0} \frac{\sin^2 x}{x^2}$ .
Find $\displaystyle\lim_{x \to \infty} \left(\sqrt{x^2 + x} - x\right)$ .
Determine whether $\displaystyle\lim_{x \to 0} \frac{1}{x^2}\sin\!\left(\frac{1}{x}\right)$ exists.

Practice Problems

Question 1: Epsilon-delta proof

Using the epsilon-delta definition, prove that $\displaystyle\lim_{x \to 2} (3x - 1) = 5$ .

Answer

We need to show: for every $\epsilon \gt 0$ There exists a $\delta \gt 0$ such that if $0 \lt |x - 2| \lt \delta$ Then $|(3x - 1) - 5| \lt \epsilon$ .

$|(3x - 1) - 5| = |3x - 6| = 3|x - 2|$ .

We need $3|x - 2| \lt \epsilon$ So $|x - 2| \lt \epsilon/3$ .

Choose $\delta = \epsilon/3$ . Then if $0 \lt |x - 2| \lt \delta$ :

$|(3x - 1) - 5| = 3|x - 2| \lt 3\delta = 3(\epsilon/3) = \epsilon$ .

Therefore, $\displaystyle\lim_{x \to 2} (3x - 1) = 5$ .

Question 2: Limits involving trigonometric functions

Evaluate $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x \sin x}$ .

Answer

Multiply numerator and denominator by $1 + \cos x$ :

$\displaystyle\lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x)}{x \sin x(1 + \cos x)} = \lim_{x \to 0} \frac{\sin^2 x}{x \sin x(1 + \cos x)}$

$= \lim_{x \to 0} \frac{\sin x}{x(1 + \cos x)} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{1 + \cos x} = 1 \cdot \frac{1}{1 + 1} = \frac{1}{2}$ .

Question 3: Continuity of a piecewise function

Determine whether the following function is continuous at $x = 1$ :

$f(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & \mathrm{if x \ne 1 \\ 4 & \mathrm{if x = 1 \end{cases}$

Answer

Check three conditions:

$f(1) = 4$ (defined).
$\displaystyle\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x + 1) = 2$ .
$\lim_{x \to 1} f(x) = 2 \ne f(1) = 4$ .

Since the limit does not equal the function value, $f$ is NOT continuous at $x = 1$ . To make it Continuous, $f(1)$ should be redefined as $2$ .

Question 4: Intermediate Value Theorem application

Prove that the equation $x^5 - 5x + 1 = 0$ has at least one root in the interval $(0, 1)$ .

Answer

Let $f(x) = x^5 - 5x + 1$ . This is a polynomial, so it is continuous everywhere.

$f(0) = 0 - 0 + 1 = 1 \gt 0$ .

$f(1) = 1 - 5 + 1 = -3 \lt 0$ .

Since $f$ is continuous on $[0, 1]$ and $f(0) \gt 0$ and $f(1) \lt 0$ By the Intermediate Value Theorem, there exists at least one $c \in (0, 1)$ such that $f(c) = 0$ .

Question 5: Squeeze theorem

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

Answer

Since $-1 \le \sin\!\left(\frac{1}{x}\right) \le 1$ for all $x \ne 0$ :

$-x^2 \le x^2 \sin\!\left(\frac{1}{x}\right) \le x^2$ .

$\displaystyle\lim_{x \to 0} (-x^2) = 0$ and $\displaystyle\lim_{x \to 0} x^2 = 0$ .

By the Squeeze Theorem: $\displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0$ .

hardest questions within the AP specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Limits and Continuity with other AP Calculus topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

flowchart TD
    A[1_Limits And Continuity] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the mathematical techniques and concepts related to limits and continuity, including key theorems, methods, and problem-solving approaches.

Key concepts include:

quadratic equations and the discriminant
simultaneous equations
polynomial division and the factor theorem
partial fractions
binomial expansion

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Intuition

The GPS navigator analogy: Limits describe where a function is heading, not where it actually arrives. Imagine driving toward a destination — limits tell you the direction and destination of your journey, even if there’s a pothole (discontinuity) at that exact spot.

Why it matters: Limits are the foundation of all calculus. Without understanding limits, derivatives (instantaneous rates of change) and integrals (accumulation of quantities) cannot be properly understood or applied.

The key insight: A function can approach a value without ever reaching it, and that approaching behaviour is what matters for calculus.

Cross-References

Derivatives — The derivative is defined as a limit of a difference quotient, making limits the foundation of differential calculus.
Integrals — The definite integral is defined as a limit of Riemann sums, connecting limits to the accumulation of quantities.
Sequences and Series — Convergence of sequences and series relies on the same limit concepts that underpin continuity.
AP Physics — Kinematics: Instantaneous velocity and acceleration are defined as limits of average velocity and acceleration over shrinking time intervals.
AP Physics — Work, Energy, and Power: The work integral is defined as a limit of Riemann sums, directly applying limit concepts to physical quantities.

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