5.2 Limits of Functions, Indeterminate Forms & Squeeze Theorem

Limits of Functions, Indeterminate Forms & Squeeze Theorem

The mathematical concept of a limit is the foundational cornerstone upon which differential calculus, integral calculus, and modern analysis are constructed. On the UET ECAT entry test, limits are heavily represented, testing both theoretical understanding (one-sided limits, Squeeze Theorem) and rapid mechanical proficiency in neutralizing indeterminate forms under severe time pressure.

---

1. The Limit Concept and One-Sided Limits

Intuitive and Formal Definition

Let $f$ be a function defined on an open interval containing $c$, except possibly at $c$ itself. We write: $\lim_{x \to c} f(x) = L$ if for every real number $\epsilon > 0$, there exists a corresponding real number $\delta > 0$ such that: $0 < |x - c| < \delta \implies |f(x) - L| < \epsilon$ This means that as $x$ approaches arbitrarily close to $c$ (from either side, without ever needing to equal $c$), $f(x)$ approaches arbitrarily close to the real finite number $L$.

Left-Hand and Right-Hand Limits

1. Left-Hand Limit (LHL): The value $f(x)$ approaches as $x$ approaches $c$ through values strictly smaller than $c$ ($x < c$): $\lim_{x \to c^-} f(x) = \lim_{h \to 0^+} f(c - h)$
2. Right-Hand Limit (RHL): The value $f(x)$ approaches as $x$ approaches $c$ through values strictly greater than $c$ ($x > c$): $\lim_{x \to c^+} f(x) = \lim_{h \to 0^+} f(c + h)$

Fundamental Existence Criterion

The two-sided limit $\lim_{x \to c} f(x)$ exists if and only if both one-sided limits exist and are finite and equal: $\lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L$

Critical ECAT Traps with One-Sided Limits

• Signum Function / Absolute Value: Consider $f(x) = \frac{|x|}{x}$ as $x \to 0$:

  • LHL: $\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1$.
  • RHL: $\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = +1$.
  • Since $\text{LHL} \neq \text{RHL}$, $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

• Exponential Fraction: Consider $f(x) = \frac{1}{1 + e^{1/x}}$ as $x \to 0$:

  • RHL ($x \to 0^+$): $\frac{1}{x} \to +\infty \implies e^{1/x} \to \infty \implies \frac{1}{1 + \infty} = 0$.
  • LHL ($x \to 0^-$): $\frac{1}{x} \to -\infty \implies e^{1/x} \to 0 \implies \frac{1}{1 + 0} = 1$.
  • Because $0 \neq 1$, the overall limit does not exist.

---

2. Limit Theorems (Algebra of Limits)

Let $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$, where $L, M \in \mathbb{R}$, and let $k$ be a scalar constant:

1. Constant Rule: $\lim_{x \to c} k = k$ and $\lim_{x \to c} x = c$.
2. Scalar Multiple Rule: $\lim_{x \to c} [k \cdot f(x)] = k \lim_{x \to c} f(x) = kL$.
3. Sum and Difference Rule: $\lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x) = L \pm M$.
4. Product Rule: $\lim_{x \to c} [f(x) \cdot g(x)] = [\lim_{x \to c} f(x)] \cdot [\lim_{x \to c} g(x)] = L \cdot M$.
5. Quotient Rule: $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)} = \frac{L}{M}$, provided $M \neq 0$.
6. Power Rule: $\lim_{x \to c} [f(x)]^n = [\lim_{x \to c} f(x)]^n = L^n$ ($n \in \mathbb{Q}$, assuming $L > 0$ for fractional $n$).

---

3. The Seven Indeterminate Forms

When direct substitution of $x = c$ into an expression produces an undefined mathematical form, the form is termed indeterminate. The seven indeterminate forms in calculus are:

$\frac{0}{0}, \quad \frac{\pm\infty}{\pm\infty}, \quad 0 \times (\pm\infty), \quad \infty - \infty, \quad 0^0, \quad 1^\infty, \quad \infty^0$

Caution: Forms such as $\frac{0}{\infty} = 0$, $\frac{\infty}{0} = \pm\infty$, $0^\infty = 0$, and $\infty + \infty = \infty$ are determinate forms and can be evaluated directly without indeterminate transformations.

Transformation Strategies for Indeterminate Forms

1. Quotient Forms ($\frac{0}{0}$ and $\frac{\infty}{\infty}$): Resolved via factoring, conjugate rationalization, dividing by dominant powers, or L'Hôpital's Rule.

2. Product Form ($0 \times \infty$): Convert to a quotient form by placing one factor in the denominator: $f(x) \cdot g(x) = \frac{f(x)}{1/g(x)} \quad \left(\frac{0}{0}\right) \qquad \text{or} \qquad \frac{g(x)}{1/f(x)} \quad \left(\frac{\infty}{\infty}\right)$

3. Difference Form ($\infty - \infty$): Typically occurs with fractions or radical differences. Combine fractions over a common algebraic denominator or rationalize via the conjugate surd.

4. Exponential Forms ($0^0, 1^\infty, \infty^0$): Let $y = [f(x)]^{g(x)}$. Take the natural logarithm of both sides: $\ln y = g(x) \ln[f(x)]$ Evaluate $L = \lim_{x \to c} \ln y$ (which is now in $0 \times \infty$ form), then exponentiate back: $\lim_{x \to c} [f(x)]^{g(x)} = e^L$

The $1^\infty$ Rapid ECAT Formula

A frequent high-yield shortcut on the ECAT: if $\lim_{x \to c} f(x) = 1$ and $\lim_{x \to c} g(x) = \pm\infty$, then: $\lim_{x \to c} [f(x)]^{g(x)} = e^{\lim_{x \to c} g(x)[f(x) - 1]}$ This formula computes the answer in a single algebraic line, completely bypassing logarithmic substitution.

---

4. Systematic Algebraic Limit Techniques

1. Factorization and the Binomial Identity

When direct substitution yields $\frac{0}{0}$ in rational expressions, the factor $(x - c)$ is present in both numerator and denominator. Factorizing and canceling $(x - c)$ resolves the indeterminate form.

Fundamental Identity: For any rational number $n$ and $a > 0$: $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$

Generalized Ratio Identity: $\lim_{x \to a} \frac{x^n - a^n}{x^m - a^m} = \frac{n}{m} a^{n - m}$

2. Conjugate Rationalization for Radical Expressions

When expressions involve $\sqrt{A} - \sqrt{B}$ yielding $\frac{0}{0}$, multiply numerator and denominator by the conjugate radical $(\sqrt{A} + \sqrt{B})$ using the difference of squares identity $(\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B}) = A - B$.

3. Limits at Infinity (Dominant Term Technique)

For rational limits as $x \to \pm\infty$: $\lim_{x \to \infty} \frac{a_n x^n + a_{n-1}x^{n-1} + \dots + a_0}{b_m x^m + b_{m-1}x^{m-1} + \dots + b_0}$ Divide every term by the highest denominator power $x^m$, or apply the Degree Comparison Rule:

• $\text{deg}(\text{Num}) < \text{deg}(\text{Den})$ ($n < m$): Limit is $0$.
• $\text{deg}(\text{Num}) = \text{deg}(\text{Den})$ ($n = m$): Limit is the ratio of leading coefficients $\frac{a_n}{b_m}$.
• $\text{deg}(\text{Num}) > \text{deg}(\text{Den})$ ($n > m$): Limit is $\pm\infty$ (no finite limit).

---

5. Standard Transcendental Limits

The Fundamental Trigonometric Limit

$\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$

Geometric Proof (Unit Circle & Sector Areas): Consider a circle of radius $r = 1$ with central angle $\theta$ ($0 < \theta < \frac{\pi}{2}$ in radians). Construct:

1. Inside triangle $OAB$ with area $\frac{1}{2}(1)(1)\sin\theta = \frac{1}{2}\sin\theta$.
2. Circular sector $OAB$ with area $\frac{1}{2}(1)^2\theta = \frac{1}{2}\theta$.
3. Outside right triangle $OAC$ with base $1$ and height $\tan\theta$, having area $\frac{1}{2}(1)\tan\theta = \frac{1}{2}\tan\theta$.

Comparing geometric areas: $\text{Area}(\triangle OAB) < \text{Area}(\text{Sector } OAB) < \text{Area}(\triangle OAC)$: $\frac{1}{2}\sin\theta < \frac{1}{2}\theta < \frac{1}{2}\tan\theta \implies \sin\theta < \theta < \frac{\sin\theta}{\cos\theta}$ Dividing by $\sin\theta > 0$ and inverting: $\cos\theta < \frac{\sin\theta}{\theta} < 1$ Taking $\lim_{\theta \to 0^+} \cos\theta = 1$, by the Squeeze Theorem, $\lim_{\theta \to 0^+} \frac{\sin\theta}{\theta} = 1$. By parity, the left-hand limit is identical, establishing the theorem.

ECAT Trap Alert — Radians vs. Degrees: The proof strictly requires $\theta$ to be in radians! If an ECAT question gives the angle in degrees: $\lim_{x \to 0} \frac{\sin x^\circ}{x} = \lim_{x \to 0} \frac{\sin\left(\frac{\pi x}{180}\right)}{x} = \frac{\pi}{180}$

Derived Trigonometric Limits

• $\lim_{x \to 0} \frac{\tan x}{x} = 1$
• $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$
• $\lim_{x \to 0} \frac{1 - \cos(kx)}{x^2} = \lim_{x \to 0} \frac{2\sin^2(kx/2)}{x^2} = 2 \left(\frac{k}{2}\right)^2 = \frac{k^2}{2}$

Euler's Number Exponential Limits

• $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e \approx 2.71828$
• $\lim_{x \to 0} (1 + x)^{1/x} = e$
• Generalized Form: $\lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^{mx} = e^{km}$
• $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \quad (a > 0)$
• When $a = e$: $\lim_{x \to 0} \frac{e^x - 1}{x} = \ln e = 1$
• $\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$

---

6. The Squeeze (Sandwich) Theorem

Theorem Statement

Let $f(x), g(x), h(x)$ be defined on an open interval containing $c$, except possibly at $c$ itself. If: $g(x) \le f(x) \le h(x)$ for all $x$ in the interval (with $x \neq c$), and: $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$ then the limit of the trapped function $f(x)$ must also exist and equal $L$: $\lim_{x \to c} f(x) = L$

Application to Oscillating Functions

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

1. Since the sine function is bounded: $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ for all $x \neq 0$.
2. Multiply throughout by $x^2 \ge 0$: $-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$
3. Take limits as $x \to 0$: $\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0$
4. By the Squeeze Theorem, $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$.

ECAT Contrast: Does $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$ exist? No. As $x \to 0$, $\frac{1}{x} \to \infty$, causing the sine curve to oscillate infinitely between $-1$ and $+1$ without approaching any single value.

---

7. L'Hôpital's Rule for High-Speed ECAT Calculations

Theorem Statement

Let $f$ and $g$ be differentiable on an open interval containing $c$ (except possibly at $c$). If direct substitution produces $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, then: $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$ provided the limit of the derivative ratio exists or is $\pm\infty$.

Crucial Rule: Differentiate the numerator $f(x)$ and denominator $g(x)$ separately. Do NOT use the quotient rule!

Speed Demonstration

Evaluate $\lim_{x \to 2} \frac{x^5 - 32}{x^3 - 8}$:

• Direct substitution gives $\frac{32 - 32}{8 - 8} = \frac{0}{0}$.
• Factoring $x^5 - 32$ takes 45 seconds of polynomial algebra.
• Applying L'Hôpital's Rule directly in 5 seconds: $\lim_{x \to 2} \frac{5x^4}{3x^2} = \frac{5(2)^4}{3(2)^2} = \frac{5(16)}{3(4)} = \frac{20}{3}$

---

8. Standard Limits Reference Table