Mathematics: Class 11 — Chapter 12: Limits and Derivatives

Consider a body dropped from a tall cliff. Physical experiments confirm that the distance $s$ (in meters) covered by the body as a function of time $t$ (in seconds) is given by:

To find the velocity of the body at exactly $t = 2$ seconds, we analyze the average velocity over smaller and smaller time intervals ending or starting at $t = 2$.

• Average Velocity between $t = t_1$ and $t = t_2$ is defined as: $$\text{Average Velocity} = \frac{\text{Distance travelled between } t_1 \text{ and } t_2}{\text{Time interval } (t_2 - t_1)} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$$

By calculating the average velocity over intervals ending at $t = 2$ (using $t_1$ values approaching 2 from the left) and starting at $t = 2$ (using $t_2$ values approaching 2 from the right), we observe the following limiting behavior:

This limiting process indicates that the instantaneous velocity at $t = 2$ is exactly 19.6 m/s. Geometrically, this corresponds to the slope of the tangent to the curve $s = 4.9t^2$ at the point $A(2, 19.6)$.

To understand limits comprehensively, we must analyze the behavior of $f(x)$ as $x$ approaches $a$ from both directions:

• Left-Hand Limit (LHL): The expected value of $f(x)$ as $x$ approaches $a$ from values strictly less than $a$. $$\lim_{x \to a^-} f(x)$$
• Right-Hand Limit (RHL): The expected value of $f(x)$ as $x$ approaches $a$ from values strictly greater than $a$. $$\lim_{x \to a^+} f(x)$$
• Existence of a Limit: The limit of $f(x)$ as $x \to a$ exists if and only if both the left-hand and right-hand limits exist and are equal: $$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = \lim_{x \to a} f(x) = l$$

Theorem 1: Let $f$ and $g$ be two functions such that both $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist. Then the following algebraic properties hold:

1. Sum Rule: The limit of a sum is the sum of the limits. $$\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$
2. Difference Rule: The limit of a difference is the difference of the limits. $$\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$$
3. Product Rule: The limit of a product is the product of the limits. $$\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$$
4. Minus / Quotient Rule: The limit of a quotient is the quotient of the limits, provided the denominator's limit is non-zero. $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \quad \text{where } \lim_{x \to a} g(x) \neq 0$$
5. Scalar Multiplication: For any real number $\lambda$: $$\lim_{x \to a} [\lambda \cdot f(x)] = \lambda \cdot \lim_{x \to a} f(x)$$

• For any polynomial function $f(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$, the limit at $x = a$ is simply the value of the polynomial at $a$: $$\lim_{x \to a} f(x) = f(a)$$
• For a rational function $f(x) = \frac{g(x)}{h(x)}$ where $g(x)$ and $h(x)$ are polynomials:
  • If $h(a) \neq 0$, then $\lim_{x \to a} f(x) = \frac{g(a)}{h(a)}$.
  • If $h(a) = 0$ and $g(a) \neq 0$, the limit does not exist.
  • If $h(a) = 0$ and $g(a) = 0$, we obtain the indeterminate form $\frac{0}{0}$. We must factor out common terms of $(x-a)$ from both numerator and denominator to evaluate the limit.

Theorem 2: For any positive integer $n$ (and indeed any rational number $n$ when $a > 0$):

Mathematical Problem Statement: Evaluate the limit: $\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$

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Step-by-Step Algorithmic Solution Trace:

1. Divide both the numerator and the denominator by $(x - 1)$ to express the limit in the standard form of Theorem 2: $$\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1} = \lim_{x \to 1} \frac{\frac{x^{15} - 1}{x - 1}}{\frac{x^{10} - 1}{x - 1}}$$
2. Apply the quotient rule of limits: $$\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1} = \frac{\lim_{x \to 1} \frac{x^{15} - 1}{x - 1}}{\lim_{x \to 1} \frac{x^{10} - 1}{x - 1}}$$
3. Apply Theorem 2 ($\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$) to both numerator ($n=15, a=1$) and denominator ($n=10, a=1$): $$\text{Numerator Limit} = 15(1)^{14} = 15$$ $$\text{Denominator Limit} = 10(1)^9 = 10$$
4. Compute the final ratio: $$\text{Limit} = \frac{15}{10} = \frac{3}{2}$$

Final Evaluation Output: $\frac{3}{2}$

Mathematical Problem Statement: Evaluate the limit: $\lim_{x \to 2} \frac{x^3 - 4x^2 + 4x}{x^2 - 4}$

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Step-by-Step Algorithmic Solution Trace:

1. Substitute $x = 2$ directly to check the form: $$\text{Numerator: } 2^3 - 4(2)^2 + 4(2) = 8 - 16 + 8 = 0$$ $$\text{Denominator: } 2^2 - 4 = 0$$ This is an indeterminate form of $\frac{0}{0}$.
2. Factorize the numerator and denominator to isolate the vanishing factor $(x - 2)$: $$\text{Numerator: } x^3 - 4x^2 + 4x = x(x^2 - 4x + 4) = x(x - 2)^2$$ $$\text{Denominator: } x^2 - 4 = (x - 2)(x + 2)$$
3. Rewrite the limit expression and cancel the common factor $(x - 2)$ (valid since $x \neq 2$): $$\lim_{x \to 2} \frac{x(x - 2)^2}{(x - 2)(x + 2)} = \lim_{x \to 2} \frac{x(x - 2)}{x + 2}$$
4. Substitute $x = 2$ into the simplified expression: $$\text{Limit} = \frac{2(2 - 2)}{2 + 2} = \frac{0}{4} = 0$$

Final Evaluation Output: $0$

Theorem 4: Let $f$, $g$, and $h$ be real-valued functions such that:

for all $x$ in a common domain of definition (except possibly at $x = a$). If:

then the limit of the middle function $g(x)$ is also forced to be $l$:

Using geometric proofs and the Sandwich Theorem, we establish the following fundamental limits (where angles are measured in radians):

Mathematical Problem Statement: Evaluate the limit: $\lim_{x \to 0} \frac{\sin 4x}{\sin 2x}$

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Step-by-Step Algorithmic Solution Trace:

1. Express the limit by introducing the appropriate scaling factors in the numerator and denominator to match the standard form $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$: $$\lim_{x \to 0} \frac{\sin 4x}{\sin 2x} = \lim_{x \to 0} \left[ \frac{\frac{\sin 4x}{4x} \cdot 4x}{\frac{\sin 2x}{2x} \cdot 2x} \right]$$
2. Simplify the algebraic terms outside the trigonometric functions: $$\lim_{x \to 0} \left[ \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4x}{2x} \right] = 2 \cdot \lim_{x \to 0} \left[ \frac{\frac{\sin 4x}{4x}}{\frac{\sin 2x}{2x}} \right]$$
3. Apply the quotient rule of limits. Note that as $x \to 0$, both $4x \to 0$ and $2x \to 0$: $$\text{Limit} = 2 \cdot \frac{\lim_{4x \to 0} \frac{\sin 4x}{4x}}{\lim_{2x \to 0} \frac{\sin 2x}{2x}}$$
4. Substitute the standard limit values: $$\text{Limit} = 2 \cdot \frac{1}{1} = 2$$

Final Evaluation Output: $2$

• Derivative at a Point: Let $f$ be a real-valued function and $a$ be a point in its domain. The derivative of $f$ at $a$, denoted by $f'(a)$, is defined as: $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ provided this limit exists.
• First Principle of Derivative: The derivative function $f'(x)$ at any general point $x$ is defined as: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Theorem 5: Let $f$ and $g$ be two functions whose derivatives are defined in a common domain. Let $u = f(x)$ and $v = g(x)$. Then:

1. Sum Rule: $$\frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)$$
2. Difference Rule: $$\frac{d}{dx}[f(x) - g(x)] = \frac{d}{dx}f(x) - \frac{d}{dx}g(x)$$
3. Product Rule (Leibnitz Rule): $$\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x) \quad \text{or} \quad (uv)' = u'v + uv'$$
4. Quotient Rule: $$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \quad \text{or} \quad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$

Mathematical Problem Statement: Find the derivative of $f(x) = \frac{1}{x}$ using the First Principle.

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Step-by-Step Algorithmic Solution Trace:

1. Set up the limit using the First Principle definition: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$$
2. Simplify the numerator by finding a common denominator: $$\frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}$$
3. Substitute this back into the limit expression: $$f'(x) = \lim_{h \to 0} \frac{\frac{-h}{x(x+h)}}{h} = \lim_{h \to 0} \left[ \frac{-h}{h \cdot x(x+h)} \right]$$
4. Cancel the common factor $h$ (valid since $h \neq 0$): $$f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)}$$
5. Evaluate the limit by direct substitution of $h = 0$: $$f'(x) = \frac{-1}{x(x+0)} = -\frac{1}{x^2}$$

Final Evaluation Output: $-\frac{1}{x^2}$

Mathematical Problem Statement: Find the derivative of $f(x) = \cot x$ using the Quotient Rule.

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Step-by-Step Algorithmic Solution Trace:

1. Express $\cot x$ as a quotient of cosine and sine: $$f(x) = \frac{\cos x}{\sin x}$$
2. Identify the numerator $u = \cos x$ and denominator $v = \sin x$. Compute their individual derivatives: $$u' = -\sin x$$ $$v' = \cos x$$
3. Apply the Quotient Rule formula $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$: $$f'(x) = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2}$$
4. Simplify the numerator using the fundamental Pythagorean identity $\sin^2 x + \cos^2 x = 1$: $$\text{Numerator} = -\sin^2 x - \cos^2 x = -(\sin^2 x + \cos^2 x) = -1$$
5. Write the final simplified expression: $$f'(x) = \frac{-1}{\sin^2 x} = -\csc^2 x$$

Final Evaluation Output: $-\csc^2 x$