Introduction
In JEE Main Mathematics, the concepts of Limits, Continuity, and Differentiability form the foundation for understanding calculus. This study note will provide a detailed explanation of these topics, breaking down complex ideas into simpler parts and using examples to illustrate key points.
Limits
Definition of a Limit
The limit of a function $f(x)$ as $x$ approaches a value $a$ is the value that $f(x)$ gets closer to as $x$ gets closer to $a$. Mathematically, it is represented as:
$$ \lim_{{x \to a}} f(x) = L $$
This means that as $x$ approaches $a$, $f(x)$ approaches $L$.
Finding Limits
There are several methods to find limits:
- Direct Substitution: If $f(x)$ is continuous at $x = a$, then $\lim_{{x \to a}} f(x) = f(a)$.
- Factoring: Factor the expression and simplify to remove the indeterminate form.
- Rationalizing: Multiply by the conjugate to simplify the expression.
- L'Hôpital's Rule: Used when the limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
Example: Find $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$.
Solution: Factor the numerator: $$ \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} $$ Cancel the common factor: $$ = x + 2 $$ Now, substitute $x = 2$: $$ \lim_{{x \to 2}} (x + 2) = 4 $$
Special Limits
- Limit at Infinity: $$\lim_{{x \to \infty}} \frac{1}{x} = 0$$
- Squeeze Theorem: If $g(x) \leq f(x) \leq h(x)$ and $\lim_{{x \to a}} g(x) = \lim_{{x \to a}} h(x) = L$, then $\lim_{{x \to a}} f(x) = L$.
Continuity
Definition of Continuity
A function $f(x)$ is continuous at a point $x = a$ if:
- $f(a)$ is defined.
- $\lim_{{x \to a}} f(x)$ exists.
- $\lim_{{x \to a}} f(x) = f(a)$.
Types of Discontinuities
- Removable Discontinuity: The limit exists, but $f(a)$ is not equal to the limit.
- Jump Discontinuity: The left-hand limit and right-hand limit exist but are not equal.
- Infinite Discontinuity: The function approaches infinity near $a$.
Example: Determine the continuity of $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.
Solution: Factor the numerator: $$ f(x) = \frac{(x - 1)(x + 1)}{x - 1} $$ Simplify: $$ f(x) = x + 1 \quad \text{for} \quad x \neq 1 $$ At $x = 1$, the function is not defined. However, the limit as $x$ approaches 1 is: $$ \lim_{{x \to 1}} (x + 1) = 2 $$ Thus, $f(x)$ has a removable discontinuity at $x = 1$.
Differentiability
Definition of Differentiability
A function $f(x)$ is differentiable at a point $x = a$ if the derivative $f'(a)$ exists. The derivative is defined as:
$$ f'(a) = \lim_{{h \to 0}} \frac{f(a + h) - f(a)}{h} $$
Relationship Between Continuity and Differentiability
- If $f(x)$ is differentiable at $x = a$, it is also continuous at $x = a$.
- However, if $f(x)$ is continuous at $x = a$, it is not necessarily differentiable at $x = a$.
A function can be continuous but not differentiable at a point (e.g., $|x|$ at $x = 0$).
Rules of Differentiation
- Power Rule: $$\frac{d}{dx} [x^n] = nx^{n-1}$$
- Product Rule: $$\frac{d}{dx} [uv] = u'v + uv'$$
- Quotient Rule: $$\frac{d}{dx} \left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$$
- Chain Rule: $$\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$$
Example: Find the derivative of $f(x) = x^3 \sin(x)$.
Solution: Use the product rule: $$ f'(x) = \frac{d}{dx} [x^3] \cdot \sin(x) + x^3 \cdot \frac{d}{dx} [\sin(x)] $$ $$ = 3x^2 \sin(x) + x^3 \cos(x) $$
Conclusion
Understanding Limits, Continuity, and Differentiability is crucial for mastering calculus. These concepts are interconnected and form the basis for more advanced topics. Practice is key to gaining proficiency, so work through various problems to solidify your understanding.
TipAlways check the conditions for continuity and differentiability. Remember that differentiability implies continuity, but the converse is not always true.
Common MistakeA common mistake is assuming that a function is differentiable at a point just because it is continuous. Always verify differentiability separately.
