Chapter 12: Limits and Derivatives
12.1 Introduction to Limits
Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. The concept of a limit is fundamental to calculus. It allows us to determine the value that a function approaches as its input approaches some value.
Let \( y = f(x) \). If \( f(x) \) gets closer and closer to a real number \( L \) as \( x \) gets closer and closer to \( a \) (from both sides), we say that the limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \), and we write: \[ \lim_{x \to a} f(x) = L \]
Algebra of Limits
Let \( \lim_{x \to a} f(x) = l \) and \( \lim_{x \to a} g(x) = m \).
- Limit of sum: \( \lim_{x \to a} [f(x) + g(x)] = l + m \)
- Limit of difference: \( \lim_{x \to a} [f(x) - g(x)] = l - m \)
- Limit of product: \( \lim_{x \to a} [f(x) \cdot g(x)] = l \cdot m \)
- Limit of quotient: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{l}{m} \), provided \( m \neq 0 \).
Standard Limits
- \[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \]
- \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \] (where x is in radians)
- \[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \]
12.2 Derivatives
Derivative of a function \( f(x) \) at a point \( a \) represents the rate of change of \( f(x) \) with respect to \( x \) at \( x = a \). It is geometrically the slope of the tangent to the curve \( y = f(x) \) at that point.
The derivative of a function \( f \) at \( a \) is defined as: \[ f’(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \] provided this limit exists.
If we view derivative as a function, then the derivative of \( f(x) \) with respect to \( x \) is: \[ f’(x) = \frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] This process of finding the derivative is called differentiation from first principles.
Algebra of Derivatives
- Sum/Difference Rule: \( \frac{d}{dx}[u \pm v] = \frac{du}{dx} \pm \frac{dv}{dx} \)
- Product Rule (Leibniz Rule): \( \frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx} \)
- Quotient Rule: \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \( v \neq 0 \).
Derivatives of Standard Functions
- \( \frac{d}{dx}(x^n) = n x^{n-1} \)
- \( \frac{d}{dx}(\sin x) = \cos x \)
- \( \frac{d}{dx}(\cos x) = -\sin x \)
- \( \frac{d}{dx}(\tan x) = \sec^2 x \)
- \( \frac{d}{dx}(\text{constant}) = 0 \)
Competency-Based Questions
Question 1:
Evaluate the limit: \( \lim_{x \to 2} \frac{x^3 - 2x^2}{x^2 - 5x + 6} \).
Answer 1:
Let \( f(x) = \frac{x^3 - 2x^2}{x^2 - 5x + 6} \).
If we substitute \( x = 2 \) directly, we get:
Numerator: \( 2^3 - 2(2^2) = 8 - 8 = 0 \)
Denominator: \( 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0 \)
This gives a \( \frac{0}{0} \) indeterminate form, which means \( (x-2) \) is a common factor in both numerator and denominator. Let’s factorize.
Numerator: \( x^3 - 2x^2 = x^2(x - 2) \)
Denominator: \( x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x(x-3) - 2(x-3) = (x-2)(x-3) \)
So, the function can be written as:
\[ \lim_{x \to 2} \frac{x^2(x - 2)}{(x-2)(x-3)} \]
Cancel the common factor \( (x-2) \) for \( x \neq 2 \):
\[ \lim_{x \to 2} \frac{x^2}{x-3} \]
Now substitute \( x = 2 \):
\[ = \frac{2^2}{2 - 3} = \frac{4}{-1} = -4 \]
Conclusion: The value of the limit is -4.
Question 2:
Find the derivative of the function \( f(x) = \frac{x + \cos x}{\tan x} \) with respect to \( x \).
Answer 2:
Here we use the Quotient Rule: \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \cdot u’ - u \cdot v’}{v^2} \).
Let \( u = x + \cos x \) and \( v = \tan x \).
Differentiate \( u \):
\[ u’ = \frac{d}{dx}(x + \cos x) = 1 - \sin x \]
Differentiate \( v \):
\[ v’ = \frac{d}{dx}(\tan x) = \sec^2 x \]
Apply the rule:
\[ f’(x) = \frac{(\tan x)(1 - \sin x) - (x + \cos x)(\sec^2 x)}{(\tan x)^2} \]
Expand the numerator:
\[ = \frac{\tan x - \tan x \sin x - x \sec^2 x - \cos x \sec^2 x}{\tan^2 x} \]
We know \( \cos x \sec^2 x = \cos x \cdot \frac{1}{\cos^2 x} = \frac{1}{\cos x} = \sec x \). And \( \tan x \sin x = \frac{\sin^2 x}{\cos x} \).
\[ f’(x) = \frac{\tan x - \frac{\sin^2 x}{\cos x} - x \sec^2 x - \sec x}{\tan^2 x} \]
Conclusion: The final derivative is \( f’(x) = \frac{\tan x - \frac{\sin^2 x}{\cos x} - x \sec^2 x - \sec x}{\tan^2 x} \).