Introduction
Limits are a fundamental concept in calculus and play a crucial role in understanding the behavior of functions as they approach specific points. In the JEE Main Mathematics syllabus, mastering limits is essential as it forms the basis for other topics such as continuity, differentiation, and integration. This study note aims to break down the concept of limits into digestible sections, providing detailed explanations, examples, and tips to help you excel in this topic.
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, this is expressed as:
$$ \lim_{{x \to a}} f(x) = L $$
where $L$ is the limit.
Formal Definition
A more rigorous definition using $\epsilon$ and $\delta$ is given by:
$$ \lim_{{x \to a}} f(x) = L \quad \text{if for every} \ \epsilon > 0, \ \text{there exists a} \ \delta > 0 \ \text{such that} \ 0
< |x - a| < \delta \ \Rightarrow \ |f(x) - L| < \epsilon $$
NoteUnderstanding this formal definition is crucial for tackling more advanced problems in limits.
One-Sided Limits
Sometimes, we are interested in the behavior of a function from one side of a point. These are called one-sided limits.
Right-Hand Limit
The right-hand limit of $f(x)$ as $x$ approaches $a$ is:
$$ \lim_{{x \to a^+}} f(x) $$
Left-Hand Limit
The left-hand limit of $f(x)$ as $x$ approaches $a$ is:
$$ \lim_{{x \to a^-}} f(x) $$
ExampleEvaluate the right-hand and left-hand limits of the function $f(x) = \frac{1}{x-1}$ as $x$ approaches 1.
- Right-hand limit: $\lim_{{x \to 1^+}} \frac{1}{x-1} = +\infty$
- Left-hand limit: $\lim_{{x \to 1^-}} \frac{1}{x-1} = -\infty$
Infinite Limits
When $f(x)$ increases or decreases without bound as $x$ approaches a certain value, we say that the limit is infinite.
Positive Infinity
$$ \lim_{{x \to a}} f(x) = +\infty $$
Negative Infinity
$$ \lim_{{x \to a}} f(x) = -\infty $$
ExampleEvaluate the limit of $f(x) = \frac{1}{x^2}$ as $x$ approaches 0.
$$ \lim_{{x \to 0}} \frac{1}{x^2} = +\infty $$
Limits at Infinity
Limits can also be evaluated as $x$ approaches infinity or negative infinity.
As $x$ Approaches Infinity
$$ \lim_{{x \to \infty}} f(x) = L $$
As $x$ Approaches Negative Infinity
$$ \lim_{{x \to -\infty}} f(x) = L $$
ExampleEvaluate the limit of $f(x) = \frac{2x^2 + 3}{x^2 - 1}$ as $x$ approaches infinity.
$$ \lim_{{x \to \infty}} \frac{2x^2 + 3}{x^2 - 1} = 2 $$
Techniques for Evaluating Limits
Direct Substitution
If $f(x)$ is continuous at $x = a$, then:
$$ \lim_{{x \to a}} f(x) = f(a) $$
Factoring
Factor the function and simplify to find the limit.
ExampleEvaluate $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$.
Factor the numerator:
$$ \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} $$
Cancel the common factor:
$$ \lim_{{x \to 2}} (x + 2) = 4 $$
Rationalizing
Rationalize the numerator or denominator to simplify the expression.
ExampleEvaluate $\lim_{{x \to 0}} \frac{\sqrt{x + 1} - 1}{x}$.
Rationalize the numerator:
$$ \frac{\sqrt{x + 1} - 1}{x} \cdot \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1} = \frac{(x + 1) - 1}{x(\sqrt{x + 1} + 1)} = \frac{x}{x(\sqrt{x + 1} + 1)} = \frac{1}{\sqrt{x + 1} + 1} $$
Now, evaluate the limit:
$$ \lim_{{x \to 0}} \frac{1}{\sqrt{x + 1} + 1} = \frac{1}{2} $$
L'Hôpital's Rule
Use L'Hôpital's Rule when you encounter indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
$$ \lim_{{x \to a}} \frac{f(x)}{g(x)} = \lim_{{x \to a}} \frac{f'(x)}{g'(x)} $$
ExampleEvaluate $\lim_{{x \to 0}} \frac{\sin x}{x}$.
Both numerator and denominator approach 0 as $x \to 0$, apply L'Hôpital's Rule:
$$ \lim_{{x \to 0}} \frac{\sin x}{x} = \lim_{{x \to 0}} \frac{\cos x}{1} = 1 $$
Important Limits
Standard Limits
- $\lim_{{x \to 0}} \frac{\sin x}{x} = 1$
- $\lim_{{x \to 0}} \frac{1 - \cos x}{x} = 0$
- $\lim_{{x \to \infty}} \left(1 + \frac{1}{x}\right)^x = e$
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:
$$ \lim_{{x \to a}} f(x) = L $$
ExampleEvaluate $\lim_{{x \to 0}} x^2 \sin \left(\frac{1}{x}\right)$.
We know that $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$, so:
$$ -x^2 \leq x^2 \sin \left(\frac{1}{x}\right) \leq x^2 $$
Applying the Squeeze Theorem:
$$ \lim_{{x \to 0}} -x^2 = 0 \quad \text{and} \quad \lim_{{x \to 0}} x^2 = 0 $$
Thus,
$$ \lim_{{x \to 0}} x^2 \sin \left(\frac{1}{x}\right) = 0 $$
Conclusion
Understanding limits is essential for mastering calculus and is a significant part of the JEE Main Mathematics syllabus. By breaking down the concepts, using examples, and applying various techniques, you can develop a strong foundation in limits. Keep practicing different types of problems to gain confidence and proficiency in this topic.
TipAlways check if direct substitution works first before applying other methods.
Common MistakeIgnoring the conditions for applying L'Hôpital's Rule can lead to incorrect results. Ensure the function is in an indeterminate form before using it.
Happy studying!
