Limits and Continuity

Why It Matters

Limits are the foundation of calculus. Every derivative, integral, and convergence proof rests on the concept of a limit. In machine learning, limits explain why gradient descent converges, why loss functions stabilize, and why certain approximations work. Without limits, you cannot understand the mathematics behind modern AI — from the backpropagation algorithm to the asymptotic analysis of training complexity.

What is a Limit?

DfLimit (Intuitive Definition)

The limit of $f(x)$ as $x$ approaches $a$ is the value $f(x)$ gets closer and closer to as $x$ gets arbitrarily close to $a$ (from either side), but not equal to $a$ . The function does not need to be defined at $a$ — only near $a$ .

Limit Notation

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

Here,

$f(x)$ =The function being evaluated
$a$ =The point x approaches
$L$ =The limit value (the value f(x) approaches)

DfLimit (Epsilon-Delta Definition)

The limit of $f(x)$ as $x$ approaches $a$ equals $L$ if for every $\epsilon > 0$ , there exists a $\delta > 0$ such that whenever $0 < |x - a| < \delta$ , it follows that $|f(x) - L| < \epsilon$ . Symbolically:

Epsilon-Delta Definition

\forall\, \epsilon > 0,\; \exists\, \delta > 0 \text{ such that } 0 < |x - a| < \delta \implies |f(x) - L| < \epsilon

Here,

$\epsilon$ =An arbitrarily small positive tolerance around L
$\delta$ =The corresponding tolerance around a
$L$ =The claimed limit value

How to think about epsilon-delta

Think of it as a challenge-response game: your opponent picks any tolerance $\epsilon$ around $L$ , and you must find a neighborhood $\delta$ around $a$ so that every $x$ in that neighborhood maps within $\epsilon$ of $L$ . If you can always win, the limit exists.

Example: Proving a Limit with Epsilon-Delta

Prove that $\lim_{x \to 2}(3x + 1) = 7$ .

We need: given $\epsilon > 0$ , find $\delta > 0$ such that $0 < |x - 2| < \delta \implies |(3x+1) - 7| < \epsilon$ .

Simplify: $|3x + 1 - 7| = |3x - 6| = 3|x - 2|$ . So we need $3|x - 2| < \epsilon$ , i.e., $|x - 2| < \epsilon/3$ .

Choose $\delta = \epsilon/3$ . Then $0 < |x - 2| < \delta \implies 3|x-2| < 3\delta = \epsilon$ . QED.

Limit Laws

ThLimit Laws (Algebraic Properties)

Assume $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ exist. Then:

Sum Law: $\lim_{x \to a}[f(x) + g(x)] = L + M$
Difference Law: $\lim_{x \to a}[f(x) - g(x)] = L - M$
Product Law: $\lim_{x \to a}[f(x) \cdot g(x)] = L \cdot M$
Quotient Law: $\lim_{x \to a}\frac{f(x)}{g(x)} = \frac{L}{M}$ , provided $M \neq 0$
Constant Multiple Law: $\lim_{x \to a}[c \cdot f(x)] = c \cdot L$ for any constant $c$
Power Law: $\lim_{x \to a}[f(x)]^n = L^n$ for any positive integer $n$
Root Law: $\lim_{x \to a}\sqrt[n]{f(x)} = \sqrt[n]{L}$ when $n$ is odd or $L > 0$
Composite Law: If $\lim_{x \to a} g(x) = L$ and $f$ is continuous at $L$ , then $\lim_{x \to a} f(g(x)) = f(L)$

ThSqueeze Theorem (Sandwich Theorem)

If $g(x) \leq f(x) \leq h(x)$ for all $x$ in an open interval containing $a$ (except possibly at $a$ ), and if $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$ , then $\lim_{x \to a} f(x) = L$ .

Example: Using the Squeeze Theorem

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

Since $-1 \leq \sin(1/x) \leq 1$ , we have $-x^2 \leq x^2 \sin(1/x) \leq x^2$ .

Both $-x^2 \to 0$ and $x^2 \to 0$ as $x \to 0$ , so by the Squeeze Theorem, $\lim_{x \to 0} x^2 \sin(1/x) = 0$ .

One-Sided Limits

DfLeft-Hand Limit

The left-hand limit of $f(x)$ as $x$ approaches $a$ is the value $f(x)$ approaches as $x$ approaches $a$ from below (values less than $a$ ):

Left-Hand Limit

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

Here,

$x \to a^-$ =x approaches a from the left (x < a)
$L$ =The left-hand limit value

DfRight-Hand Limit

The right-hand limit of $f(x)$ as $x$ approaches $a$ is the value $f(x)$ approaches as $x$ approaches $a$ from above (values greater than $a$ ):

Right-Hand Limit

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

Here,

$x \to a^+$ =x approaches a from the right (x > a)
$L$ =The right-hand limit value

ThTwo-Sided Limit Exists iff Both One-Sided Limits Exist and Are Equal

$\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$ .

Example: One-Sided Limits with the Sign Function

Consider $f(x) = \text{sgn}(x) = \begin{cases} -1 & x < 0 \\ 0 & x = 0 \\ 1 & x > 0 \end{cases}$ .

$\lim_{x \to 0^-} \text{sgn}(x) = -1$
$\lim_{x \to 0^+} \text{sgn}(x) = 1$

Since $-1 \neq 1$ , the two-sided limit $\lim_{x \to 0} \text{sgn}(x)$ does not exist.

Infinite Limits and Limits at Infinity

DfInfinite Limit

We write $\lim_{x \to a} f(x) = \infty$ if $f(x)$ grows without bound as $x$ approaches $a$ . This does not mean the limit exists in the traditional sense — it is a way of describing the behavior of unbounded growth.

DfLimit at Infinity

We write $\lim_{x \to \infty} f(x) = L$ if $f(x)$ approaches a finite value $L$ as $x$ grows without bound. This means $f$ has a horizontal asymptote $y = L$ .

Limits at Infinity for Rational Functions

\lim_{x \to \infty} \frac{a_n x^n + \cdots}{b_m x^m + \cdots} = \begin{cases} 0 & \text{if } n < m \\ \frac{a_n}{b_m} & \text{if } n = m \\ \pm\infty & \text{if } n > m \end{cases}

Here,

$n$ =Degree of the numerator
$m$ =Degree of the denominator
$a_n, b_m$ =Leading coefficients

Horizontal Asymptotes

To find horizontal asymptotes, compute $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$ . If either limit is finite, that value defines a horizontal asymptote.

Common Limits

Limit	Value	Context
$\lim_{x \to 0}\frac{\sin x}{x}$	$1$	Fundamental trigonometric limit
$\lim_{x \to 0}\frac{1 - \cos x}{x}$	$0$	Follows from $\sin^2 + \cos^2 = 1$
$\lim_{x \to 0}\frac{e^x - 1}{x}$	$1$	Definition of $e^x$ derivative at 0
$\lim_{x \to 0}\frac{\ln(1+x)}{x}$	$1$	Definition of $\ln$ derivative at 0
$\lim_{x \to 0}\frac{(1+x)^r - 1}{x}$	$r$	Generalized binomial limit
$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$	$e$	Definition of Euler's number
$\lim_{n \to \infty}\left(1 + \frac{k}{n}\right)^n$	$e^k$	Generalized exponential limit
$\lim_{x \to 0}\frac{\tan x}{x}$	$1$	Since $\tan x = \sin x / \cos x$
$\lim_{x \to 0}\frac{\arcsin x}{x}$	$1$	Inverse trigonometric limit
$\lim_{x \to 0}\frac{\arctan x}{x}$	$1$	Inverse trigonometric limit
$\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^x$	$e$	Continuous compounding
$\lim_{x \to 0}\frac{a^x - 1}{x}$	$\ln a$	General exponential limit

Squeeze Theorem

ThSqueeze Theorem (Formal Statement)

Let $f$ , $g$ , and $h$ be functions defined on an open interval containing $a$ , except possibly at $a$ itself. If $g(x) \leq f(x) \leq h(x)$ for all $x$ in the interval and

Squeeze Theorem Condition

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

Here,

$g(x)$ =Lower bound function
$h(x)$ =Upper bound function
$f(x)$ =The function sandwiched between them
$L$ =The common limit of the bounding functions

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

Example: Squeeze Theorem Application

Find $\lim_{x \to 0} \frac{\sin(x^2)}{x}$ .

Multiply top and bottom by $x$ : $\frac{\sin(x^2)}{x} = x \cdot \frac{\sin(x^2)}{x^2}$ .

As $x \to 0$ , we have $x^2 \to 0$ , so $\frac{\sin(x^2)}{x^2} \to 1$ .

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

L'Hôpital's Rule

ThL'Hôpital's Rule

If $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ (both approach 0) or both approach $\pm\infty$ , and if $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists (or equals $\pm\infty$ ), then:

L'Hôpital's Rule

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

Here,

$f(x)$ =Numerator (approaches 0 or ±∞)
$g(x)$ =Denominator (approaches 0 or ±∞)
$f'(x)$ =Derivative of numerator
$g'(x)$ =Derivative of denominator

When NOT to use L'Hôpital's Rule

L'Hôpital's Rule only applies to indeterminate forms $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ . If substituting $x = a$ gives a form like $\frac{0}{5}$ or $\frac{3}{0}$ , do not use L'Hôpital's Rule — evaluate directly.

Example: L'Hôpital's Rule

Compute $\lim_{x \to 0}\frac{e^x - 1 - x}{x^2}$ .

Substituting $x = 0$ : $\frac{e^0 - 1 - 0}{0} = \frac{0}{0}$ — indeterminate.

Apply L'Hôpital: $\lim_{x \to 0}\frac{e^x - 1}{2x} = \frac{0}{0}$ again.

Apply L'Hôpital again: $\lim_{x \to 0}\frac{e^x}{2} = \frac{1}{2}$ .

Example: L'Hôpital's Rule — Infinite Form

Compute $\lim_{x \to \infty}\frac{x^2}{e^x}$ .

Substituting $x = \infty$ : $\frac{\infty}{\infty}$ — indeterminate.

Apply L'Hôpital: $\lim_{x \to \infty}\frac{2x}{e^x} = \frac{\infty}{\infty}$ .

Apply L'Hôpital again: $\lim_{x \to \infty}\frac{2}{e^x} = 0$ .

This shows that exponentials grow faster than polynomials.

Continuity

DfContinuity at a Point

A function $f$ is continuous at $x = a$ if and only if all three conditions hold:

$f(a)$ is defined (the function exists at $a$ )
$\lim_{x \to a} f(x)$ exists (the left and right limits are equal)
$\lim_{x \to a} f(x) = f(a)$ (the limit equals the function value)

DfContinuity on an Interval

A function is continuous on an interval if it is continuous at every point in the interval. Continuous functions have no breaks, jumps, or holes.

DfTypes of Discontinuities

Removable Discontinuity: The limit exists but $f(a)$ is either undefined or not equal to the limit. You can "fix" it by redefining $f(a)$ .

Jump Discontinuity: The left and right limits both exist but are not equal. The function "jumps" from one value to another.

Infinite Discontinuity: At least one of the one-sided limits is $\pm\infty$ . The function has a vertical asymptote.

Oscillating Discontinuity: The function oscillates infinitely (e.g., $\sin(1/x)$ near $x = 0$ ), so the limit does not exist.

Type	Condition	Example	Fixable?
Removable	$\lim_{x \to a} f(x) = L$ but $f(a) \neq L$ or $f(a)$ undefined	$f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$	Yes (redefine $f(a)$ )
Jump	$\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$	$f(x) = \lfloor x \rfloor$ at integers	No
Infinite	$\lim_{x \to a} f(x) = \pm\infty$	$f(x) = \frac{1}{x}$ at $x=0$	No
Oscillating	Limit does not exist due to oscillation	$f(x) = \sin(1/x)$ at $x=0$	No

Intermediate Value Theorem

ThIntermediate Value Theorem (IVT)

If $f$ is continuous on the closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$ , then there exists at least one $c \in (a, b)$ such that $f(c) = N$ .

Intuition

A continuous function cannot "skip" values. If you draw it without lifting your pen, and it goes from $f(a) = 2$ to $f(b) = 5$ , it must pass through every value between 2 and 5.

Example: Applying the IVT

Show that $f(x) = x^3 - x - 1$ has a root in $[1, 2]$ .

$f(1) = 1 - 1 - 1 = -1 < 0$
$f(2) = 8 - 2 - 1 = 5 > 0$

Since $f$ is continuous (polynomial) and $0$ is between $f(1)$ and $f(2)$ , by IVT there exists $c \in (1, 2)$ such that $f(c) = 0$ .

Limits and Derivatives

DfThe Derivative as a Limit

The derivative of $f$ at $x$ is defined as a limit of the difference quotient:

Derivative Definition via Limits

f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

Here,

$f'(x)$ =The derivative (instantaneous rate of change)
$h$ =The increment (approaches 0)
$\frac{f(x+h) - f(x)}{h}$ =The difference quotient (average rate of change)

Key Connection

The entire theory of derivatives rests on limits. Without limits, the concept of an "instantaneous rate of change" is undefined. This is why limits are taught before derivatives — they are the rigorous foundation beneath calculus.

Python Implementation: Numerical Verification of Limits

import numpy as np

def verify_limit(func, a, expected, approach='both', tol=1e-8):
    """Numerically verify that lim_{x -> a} f(x) = expected."""
    results = {}

    if approach in ('both', 'left'):
        x_left = a - np.array([1e-1, 1e-2, 1e-3, 1e-4, 1e-5, 1e-6])
        x_left = x_left[x_left != a]  # exclude a itself
        vals_left = [func(x) for x in x_left]
        results['left'] = vals_left

    if approach in ('both', 'right'):
        x_right = a + np.array([1e-1, 1e-2, 1e-3, 1e-4, 1e-5, 1e-6])
        vals_right = [func(x) for x in x_right]
        results['right'] = vals_right

    if approach == 'both':
        final = (np.mean(vals_left[-2:]) + np.mean(vals_right[-2:])) / 2
    elif approach == 'left':
        final = np.mean(vals_left[-2:])
    else:
        final = np.mean(vals_right[-2:])

    return final, abs(final - expected) < tol


# Example 1: sin(x)/x -> 1
val, ok = verify_limit(lambda x: np.sin(x) / x if x != 0 else 1.0, a=0, expected=1.0)
print(f"lim x->0 sin(x)/x = {val:.10f}  (expected 1.0)  {'PASS' if ok else 'FAIL'}")

# Example 2: (e^x - 1)/x -> 1
val, ok = verify_limit(lambda x: (np.exp(x) - 1) / x if x != 0 else 1.0, a=0, expected=1.0)
print(f"lim x->0 (e^x-1)/x = {val:.10f}  (expected 1.0)  {'PASS' if ok else 'FAIL'}")

# Example 3: (1 + 1/n)^n -> e
def compound(n):
    return (1 + 1/n) ** n if n != 0 else np.e

val, ok = verify_limit(compound, a=10000, expected=np.e, approach='right')
print(f"lim n->inf (1+1/n)^n = {val:.10f}  (expected {np.e:.10f})  {'PASS' if ok else 'FAIL'}")

# Example 4: Numerical derivative (limit definition)
def f(x):
    return x**2

h_values = np.array([1e-1, 1e-2, 1e-3, 1e-4, 1e-5])
numerical_derivs = [(f(1 + h) - f(1)) / h for h in h_values]
print(f"\nNumerical derivative of x^2 at x=1:")
for h, d in zip(h_values, numerical_derivs):
    print(f"  h={h:.0e}: {d:.10f}")
print(f"  Expected: 2.0")

# Example 5: Squeeze theorem verification — x^2 sin(1/x)
def squeeze_func(x):
    return x**2 * np.sin(1/x) if x != 0 else 0.0

val, ok = verify_limit(squeeze_func, a=0, expected=0.0)
print(f"\nlim x->0 x^2*sin(1/x) = {val:.10f}  (expected 0.0)  {'PASS' if ok else 'FAIL'}")

Applications in AI/ML

Gradient Descent Convergence

Limits in Optimization

When we say gradient descent "converges," we mean $\lim_{k \to \infty} \nabla L(\theta_k) = 0$ . The learning rate $\eta$ must satisfy $\eta \to 0$ or follow specific schedules so that the limit exists. Without limits, we cannot prove convergence guarantees.

Asymptotic Analysis

Limits allow us to compare algorithm complexity:

$O(n)$ vs $O(n \log n)$ : we evaluate $\lim_{n \to \infty} \frac{n \log n}{n} = \lim_{n \to \infty} \log n = \infty$ , confirming $n \log n$ is asymptotically larger.
Training time for large models: $\lim_{N \to \infty} \frac{\text{FLOPS}}{N}$ determines cost scaling.

Convergence of Loss Functions

Convergence Condition

\lim_{t \to \infty} \mathcal{L}(\theta_t) = \mathcal{L}^*

Here,

$\mathcal{L}(\theta_t)$ =Loss at training step t
$\mathcal{L}^*$ =Optimal (minimum) loss

Probability and Statistics

Law of Large Numbers: $\lim_{n \to \infty} \bar{X}_n = \mu$ (sample mean converges to population mean)
Central Limit Theorem: Distributional limits underpin confidence intervals
Bayesian posterior: $\lim_{n \to \infty} P(\theta | x_1, \ldots, x_n) = \delta_{\theta^*}$ (posterior concentrates at true parameter)

Neural Network Expressiveness

Universal Approximation

The Universal Approximation Theorem states that for any continuous function $f$ on $[0,1]^n$ and any $\epsilon > 0$ , there exists a neural network $g$ such that $\sup_{x} |f(x) - g(x)| < \epsilon$ . This is fundamentally an epsilon-type statement about the limit of approximation quality as network width grows.

Common Mistakes

Mistake	Why It's Wrong	Correct Approach
Assuming $\lim_{x \to a} f(x) = f(a)$ always	Only true for continuous functions	Check all three continuity conditions
Applying L'Hôpital to non-indeterminate forms	$\frac{5}{0}$ is not $\frac{0}{0}$	Evaluate directly; only use L'Hôpital for $\frac{0}{0}$ or $\frac{\infty}{\infty}$
Thinking $\infty$ is a number	$\infty$ is a concept, not a value	Use limits to describe unbounded behavior rigorously
Confusing limit existence with limit value	The limit can exist and equal a finite value, or not exist	Check both sides: left = right?
Forgetting to check $\delta$ exists for all $\epsilon$	Epsilon-delta requires universal quantification	Verify for every $\epsilon > 0$ , not just small ones
Assuming $\lim f(x) \cdot g(x) = \lim f(x) \cdot \lim g(x)$ always	Only valid when both limits exist (finite)	Check existence first; $\infty \cdot 0$ is indeterminate
Canceling $x$ in $\frac{x}{x^2}$ without care	Must account for domain ( $x \neq 0$ )	Factor and cancel before taking the limit

Interview Questions

Q1: What is the epsilon-delta definition of a limit, and why is it needed?

A: The epsilon-delta definition provides a rigorous foundation for limits. It eliminates ambiguity by formalizing "approaches" with precise tolerances. Intuitive notions of limits fail for pathological functions (like $\sin(1/x)$ near 0). The epsilon-delta definition is needed to prove limit laws, establish the correctness of L'Hôpital's Rule, and build calculus on solid logical foundations.

Q2: When does $\lim_{x \to a} \frac{f(x)}{g(x)}$ exist even though direct substitution fails?

A: When both $f(a) = 0$ and $g(a) = 0$ (or both $\pm\infty$ ), we have an indeterminate form. The limit may still exist. Techniques include:

Factor and cancel common factors
Apply L'Hôpital's Rule (differentiate top and bottom)
Use series expansion (Taylor series)
Use the Squeeze Theorem

Q3: Explain the relationship between one-sided limits and continuity.

A: A function is continuous at $a$ if and only if:

The left-hand limit $\lim_{x \to a^-} f(x)$ exists
The right-hand limit $\lim_{x \to a^+} f(x)$ exists
Both are equal to $f(a)$

If the one-sided limits exist but differ, there is a jump discontinuity. If one or both don't exist, continuity fails.

Q4: Why can't we just use L'Hôpital's Rule for every limit problem?

A: L'Hôpital's Rule only applies to indeterminate forms ( $\frac{0}{0}$ or $\frac{\infty}{\infty}$ ). Applying it to non-indeterminate forms gives incorrect results. For example, $\lim_{x \to 1} \frac{x + 1}{x} = \frac{2}{1} = 2$ , but applying L'Hôpital gives $\frac{1}{1} = 1$ , which is wrong. Additionally, L'Hôpital requires the derivatives to exist and the limit of the ratio of derivatives to exist.

Q5: How does the Squeeze Theorem help in machine learning?

A: The Squeeze Theorem is used to:

Prove convergence of algorithms when direct evaluation is difficult
Establish bounds on error terms in numerical methods
Show that noise terms vanish: if you can bound the noise between two functions that both go to 0, the noise itself vanishes
Prove that regularized loss functions converge to their unregularized counterparts as the regularization parameter goes to 0

Q6: What happens when you take the limit of a sequence of functions? Is it always continuous?

A: No. The limit of continuous functions can be discontinuous. Consider $f_n(x) = x^n$ on $[0, 1]$ : each $f_n$ is continuous, but $\lim_{n \to \infty} f_n(x)$ equals 0 for $x < 1$ and 1 at $x = 1$ , which is discontinuous. Uniform convergence (a stronger condition than pointwise convergence) preserves continuity.

Practice Problems

Problem 1: Compute the Limit

Evaluate $\lim_{x \to 0}\frac{\sqrt{1 + x} - 1}{x}$ .

Solution

Multiply by the conjugate:

\frac{\sqrt{1+x} - 1}{x} \cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} = \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \frac{x}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}

\lim_{x \to 0} \frac{1}{\sqrt{1+x}+1} = \frac{1}{\sqrt{1}+1} = \frac{1}{2}

Problem 2: Continuity Analysis

Determine where $f(x) = \frac{x^2 - 4}{x^2 - 5x + 6}$ is discontinuous and classify each.

Solution

Factor: $f(x) = \frac{(x-2)(x+2)}{(x-2)(x-3)}$ .

At $x = 2$ : removable discontinuity. $\lim_{x \to 2} f(x) = \frac{x+2}{x-3}\Big|_{x=2} = \frac{4}{-1} = -4$ . Redefine $f(2) = -4$ to fix.
At $x = 3$ : infinite discontinuity. $\lim_{x \to 3} f(x) = \frac{5}{0} = \pm\infty$ . Vertical asymptote at $x = 3$ .

Problem 3: L'Hôpital's Rule

Evaluate $\lim_{x \to 0}\frac{\sin x - x}{x^3}$ .

Solution

Substituting gives $\frac{0}{0}$ . Apply L'Hôpital three times:

$\lim_{x \to 0}\frac{\cos x - 1}{3x^2} = \frac{0}{0}$
$\lim_{x \to 0}\frac{-\sin x}{6x} = \frac{0}{0}$
$\lim_{x \to 0}\frac{-\cos x}{6} = \frac{-1}{6}$

Answer: $-\frac{1}{6}$ .

Problem 4: Using the Squeeze Theorem

Prove that $\lim_{x \to 0} x \cos\!\left(\frac{1}{x}\right) = 0$ .

Solution

Since $-1 \leq \cos(1/x) \leq 1$ , we have $-|x| \leq x \cos(1/x) \leq |x|$ .

As $x \to 0$ : $\lim_{x \to 0}(-|x|) = 0$ and $\lim_{x \to 0}|x| = 0$ .

By the Squeeze Theorem, $\lim_{x \to 0} x \cos(1/x) = 0$ .

Problem 5: Limit at Infinity

Evaluate $\lim_{x \to \infty}\frac{3x^2 + 2x}{5x^2 - 1}$ .

Solution

Divide numerator and denominator by $x^2$ :

\lim_{x \to \infty}\frac{3 + \frac{2}{x}}{5 - \frac{1}{x^2}} = \frac{3 + 0}{5 - 0} = \frac{3}{5}

Since the degrees of numerator and denominator are equal, the limit is the ratio of leading coefficients: $\frac{3}{5}$ .

Quick Reference

Concept	Formula / Rule	Key Point
Limit Definition	$\lim_{x \to a} f(x) = L$	$f(x) \to L$ as $x \to a$
Epsilon-Delta	$\forall \epsilon>0, \exists \delta>0: 0<\|x-a\|<\delta \Rightarrow \|f(x)-L\|<\epsilon$	Rigorous definition
Sum Law	$\lim[f+g] = \lim f + \lim g$	Requires both limits exist
Product Law	$\lim[fg] = \lim f \cdot \lim g$	Requires both limits exist
Quotient Law	$\lim[f/g] = \lim f / \lim g$	Denominator limit $\neq 0$
Squeeze Theorem	$g \leq f \leq h$ , $\lim g = \lim h = L$ $\Rightarrow$ $\lim f = L$	Sandwich $f$ between bounds
L'Hôpital's Rule	$\lim \frac{f}{g} = \lim \frac{f'}{g'}$	Only for $\frac{0}{0}$ or $\frac{\infty}{\infty}$
Continuity	$\lim_{x \to a} f(x) = f(a)$	No breaks, jumps, or holes
IVT	$f$ continuous on $[a,b]$ , $N$ between $f(a),f(b)$ $\Rightarrow$ $\exists c: f(c)=N$	Continuous functions don't skip values
Derivative	$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$	Derivative is a limit
$\lim_{x \to 0}\frac{\sin x}{x} = 1$	Fundamental trig limit	Foundation for derivatives of trig functions
$\lim_{n \to \infty}(1+1/n)^n = e$	Definition of $e$	Compound interest, exponential growth

Cross-References

025 — Derivatives and Differentiation: The derivative is defined as a limit of the difference quotient.
026 — Chain Rule: Chain rule relies on limits of composite functions.
027 — Partial Derivatives: Multivariable limits and partial derivatives.
028 — Integrals: Integrals are defined as limits of Riemann sums.
029 — Taylor Series: Taylor series convergence depends on limit behavior.
030 — Optimization: Optimization requires limits to define derivatives and find critical points.
062 — Gradient Descent: Convergence proofs use limit theory.
042 — Central Limit Theorem: Distributional limits in probability.

Limits and Continuity