Welcome, future engineers, to a deep dive into one of the most fundamental and powerful concepts in calculus:
Limits. In our previous discussions, we built an intuition for what a limit means β the value a function "approaches" as its input "approaches" some point. Now, it's time to solidify that understanding with rigor, formal definitions, advanced properties, and crucial applications for JEE.
Think of limits as the bedrock upon which continuity, differentiability, and integration are built. Mastering limits isn't just about solving problems; it's about developing a profound understanding of how functions behave.
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### 1. The Formal Definition of a Limit: Epsilon-Delta ($epsilon$-$delta$)
While our intuitive understanding of limits is excellent for getting started, mathematics demands precision. What does "approaches" truly mean? This is where the
epsilon-delta definition comes in. This definition is a cornerstone of real analysis and is particularly important for JEE Advanced level conceptual understanding, though direct $epsilon$-$delta$ proofs are less common in Mains.
Let's break it down:
Definition: A function $f(x)$ has a limit $L$ as $x$ approaches $a$, written as $lim_{x o a} f(x) = L$, if for every number $mathbf{epsilon > 0}$ (epsilon), there exists a corresponding number $mathbf{delta > 0}$ (delta) such that if $mathbf{0 < |x - a| < delta}$, then $mathbf{|f(x) - L| < epsilon}$.
Let's unpack this:
*
$epsilon > 0$ (epsilon): This represents a tiny, arbitrary positive number. Think of it as a "tolerance" level for how close $f(x)$ must be to $L$. It dictates the desired accuracy for $f(x)$. The "for every $epsilon > 0$" part means we must be able to satisfy the condition no matter how small an error tolerance $epsilon$ is chosen.
*
$delta > 0$ (delta): This is another tiny positive number, which typically depends on $epsilon$. It represents how close $x$ must be to $a$ to guarantee that $f(x)$ is within $epsilon$ of $L$.
*
$0 < |x - a| < delta$: This inequality means $x$ is within a distance of $delta$ from $a$, but $x$ is *not equal to* $a$. This beautifully captures the idea of "approaching" without necessarily "reaching" $a$.
*
$|f(x) - L| < epsilon$: This means the value of $f(x)$ is within a distance of $epsilon$ from $L$.
Intuition: Imagine $L$ is a target, and $epsilon$ is the radius of a small target circle around $L$. The definition says that no matter how small you make that target circle (how tiny $epsilon$ is), I can always find a small enough interval around $a$ (with radius $delta$) such that any $x$ value from that interval (excluding $a$ itself) will produce an $f(x)$ that falls inside your target circle around $L$.
JEE Advanced Focus: While you might not be asked to prove limits using $epsilon$-$delta$ directly in a typical JEE Mains paper, understanding this definition enhances your conceptual grip on limits, continuity, and differentiability, which is crucial for advanced problems.
Example 1: Proving a Limit using $epsilon$-$delta$ definition
Let's prove that $lim_{x o 2} (3x+1) = 7$.
Proof:
1. We are given $f(x) = 3x+1$, $a=2$, and $L=7$.
2. We need to show that for any $epsilon > 0$, there exists a $delta > 0$ such that if $0 < |x - 2| < delta$, then $|(3x+1) - 7| < epsilon$.
3. Let's start with the conclusion and work backwards to find $delta$ in terms of $epsilon$:
$|(3x+1) - 7| < epsilon$
$|3x - 6| < epsilon$
$|3(x - 2)| < epsilon$
$3|x - 2| < epsilon$
$|x - 2| < frac{epsilon}{3}$
4. Comparing this with our condition $0 < |x - 2| < delta$, we can choose $delta = frac{epsilon}{3}$.
5.
Conclusion: For any given $epsilon > 0$, if we choose $delta = frac{epsilon}{3}$, then whenever $0 < |x - 2| < delta$, we have $0 < |x - 2| < frac{epsilon}{3}$, which implies $3|x - 2| < epsilon$, and thus $|3(x - 2)| < epsilon$, leading to $|(3x+1) - 7| < epsilon$.
Hence, $lim_{x o 2} (3x+1) = 7$ is proven.
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### 2. Left Hand Limit (LHL) and Right Hand Limit (RHL)
A limit exists at a point only if the function approaches the *same value* regardless of whether you approach the point from the left side or the right side.
*
Left Hand Limit (LHL): This is the value $f(x)$ approaches as $x$ approaches $a$ from values *less than* $a$. We denote it as $lim_{x o a^-} f(x)$.
*
Right Hand Limit (RHL): This is the value $f(x)$ approaches as $x$ approaches $a$ from values *greater than* $a$. We denote it as $lim_{x o a^+} f(x)$.
Condition for Existence of a Limit:
For $lim_{x o a} f(x)$ to exist, it is necessary and sufficient that:
1. $lim_{x o a^-} f(x)$ exists.
2. $lim_{x o a^+} f(x)$ exists.
3.
$lim_{x o a^-} f(x) = lim_{x o a^+} f(x) = L$ (a finite value).
Graphical Interpretation: If you can trace the graph of the function towards $x=a$ from both the left and the right, and your pen ends up at the same height (y-value $L$) in both cases, then the limit exists. If the pen ends up at different heights, the limit does not exist.
Example 2: LHL and RHL in action
Consider the function $f(x) = egin{cases} x+2 & ext{if } x < 1 \ x^2 & ext{if } x ge 1 end{cases}$.
Let's find $lim_{x o 1} f(x)$.
1.
LHL at $x=1$:
$lim_{x o 1^-} f(x) = lim_{x o 1^-} (x+2)$ (since $x < 1$)
As $x$ approaches 1 from the left, $x+2$ approaches $1+2 = 3$.
So, LHL = 3.
2.
RHL at $x=1$:
$lim_{x o 1^+} f(x) = lim_{x o 1^+} (x^2)$ (since $x ge 1$)
As $x$ approaches 1 from the right, $x^2$ approaches $1^2 = 1$.
So, RHL = 1.
Since LHL (3) $
e$ RHL (1), the limit $lim_{x o 1} f(x)$
does not exist.
---
### 3. Algebra of Limits (Properties of Limits)
Limits behave very nicely with arithmetic operations. If $lim_{x o a} f(x) = L$ and $lim_{x o a} g(x) = M$, and $c$ is a constant, then:
Property |
Description |
Formula |
|---|
Sum Rule |
The limit of a sum is the sum of the limits. |
$lim_{x o a} [f(x) + g(x)] = L + M$ |
Difference Rule |
The limit of a difference is the difference of the limits. |
$lim_{x o a} [f(x) - g(x)] = L - M$ |
Constant Multiple Rule |
The limit of a constant times a function is the constant times the limit of the function. |
$lim_{x o a} [c cdot f(x)] = c cdot L$ |
Product Rule |
The limit of a product is the product of the limits. |
$lim_{x o a} [f(x) cdot g(x)] = L cdot M$ |
Quotient Rule |
The limit of a quotient is the quotient of the limits (provided the limit of the denominator is not zero). |
$lim_{x o a} frac{f(x)}{g(x)} = frac{L}{M}$, provided $M
e 0$ |
Power Rule |
The limit of a function raised to a power is the limit of the function raised to that power. |
$lim_{x o a} [f(x)]^n = L^n$, for any integer $n$. |
Root Rule |
The limit of the nth root of a function is the nth root of the limit of the function. |
$lim_{x o a} sqrt[n]{f(x)} = sqrt[n]{L}$, provided $sqrt[n]{L}$ is real. |
Important Note (JEE Focus): These properties are valid
only if the individual limits $L$ and $M$ exist and are finite. If you encounter expressions like $L/0$ where $L
e 0$, the limit is $pm infty$ or does not exist. If it's $0/0$, $infty/infty$, etc., we call them indeterminate forms, which require further analysis.
---
### 4. Indeterminate Forms and Methods of Evaluation
When direct substitution of $x=a$ into $f(x)$ results in an expression that doesn't give a definite value, we call it an
indeterminate form. These forms do not tell us the limit directly, but signal that further simplification or transformation of the function is needed.
The seven common indeterminate forms are:
1. $0/0$
2. $infty/infty$
3. $infty - infty$
4. $0 imes infty$
5. $1^infty$
6. $0^0$
7. $infty^0$
Methods to Resolve Indeterminate Forms:
1.
Factorization and Cancellation (for $0/0$): If the numerator and denominator both become zero, it means $(x-a)$ is a common factor. Factor it out and cancel.
Example 3: $lim_{x o 2} frac{x^2 - 4}{x - 2}$
Direct substitution gives $0/0$.
$lim_{x o 2} frac{(x-2)(x+2)}{x - 2} = lim_{x o 2} (x+2)$ (since $x
e 2$, we can cancel $x-2$)
$= 2+2 = 4$.
2.
Rationalization (for $0/0$ involving square roots): Multiply the numerator and denominator by the conjugate of the expression involving roots.
Example 4: $lim_{x o 0} frac{sqrt{1+x} - 1}{x}$
Direct substitution gives $0/0$.
$lim_{x o 0} frac{sqrt{1+x} - 1}{x} imes frac{sqrt{1+x} + 1}{sqrt{1+x} + 1}$
$= lim_{x o 0} frac{(1+x) - 1}{x(sqrt{1+x} + 1)} = lim_{x o 0} frac{x}{x(sqrt{1+x} + 1)}$
$= lim_{x o 0} frac{1}{sqrt{1+x} + 1}$ (since $x
e 0$, we can cancel $x$)
$= frac{1}{sqrt{1+0} + 1} = frac{1}{1+1} = frac{1}{2}$.
3.
Using Standard Limits: Many limits involving trigonometric, exponential, or logarithmic functions can be resolved by transforming them into standard forms.
4.
L'Hopital's Rule: This is a powerful tool for $0/0$ and $infty/infty$ forms.
Note: This rule involves derivatives and will be covered in detail in the "Differentiability" section. For now, focus on algebraic manipulation and standard limits.
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### 5. Important Standard Limits (JEE Essentials)
These limits are frequently used in JEE problems and should be memorized and understood. Some derivations are crucial for a deep understanding.
A. Algebraic Standard Limit:
* $lim_{x o a} frac{x^n - a^n}{x - a} = n a^{n-1}$ (for any rational $n$)
Derivation (for integer $n$):
We know that $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + dots + xa^{n-2} + a^{n-1})$.
So, $lim_{x o a} frac{x^n - a^n}{x - a} = lim_{x o a} (x^{n-1} + x^{n-2}a + dots + a^{n-1})$
As $x o a$, each term becomes $a^{n-1}$. There are $n$ such terms.
$= a^{n-1} + a^{n-1} + dots + a^{n-1}$ ($n$ times)
$= n a^{n-1}$.
B. Trigonometric Standard Limits:
These limits are foundational. We'll derive the first one using a geometric argument.
*
$lim_{x o 0} frac{sin x}{x} = 1$ (where $x$ is in radians)
Derivation using Sandwich Theorem (Geometric Proof):
Consider a circle of radius 1 (unit circle) with center O. Let $angle AOP = x$ radians, where $x in (0, pi/2)$.
Draw $PM perp OA$. Extend OP to meet the tangent at A at point T.
We can see that:
Area of $ riangle OAP <$ Area of Sector $OAP <$ Area of $ riangle OAT$
1. Area of $ riangle OAP = frac{1}{2} cdot OA cdot PM = frac{1}{2} cdot 1 cdot sin x = frac{1}{2} sin x$.
2. Area of Sector $OAP = frac{1}{2} r^2 x = frac{1}{2} (1)^2 x = frac{1}{2} x$.
3. Area of $ riangle OAT = frac{1}{2} cdot OA cdot AT = frac{1}{2} cdot 1 cdot an x = frac{1}{2} an x$.
So, $frac{1}{2} sin x < frac{1}{2} x < frac{1}{2} an x$.
Multiplying by 2: $sin x < x < an x$.
Since $x in (0, pi/2)$, $sin x > 0$. Divide by $sin x$:
$1 < frac{x}{sin x} < frac{ an x}{sin x} Rightarrow 1 < frac{x}{sin x} < frac{1}{cos x}$.
Taking reciprocals (and reversing inequalities because all terms are positive):
$cos x < frac{sin x}{x} < 1$.
Now, as $x o 0$, we know $lim_{x o 0} cos x = cos(0) = 1$.
Also, $lim_{x o 0} 1 = 1$.
By the
Sandwich Theorem (or Squeeze Play Theorem), since $frac{sin x}{x}$ is "sandwiched" between $cos x$ and 1, and both $cos x$ and 1 approach 1 as $x o 0$, then $lim_{x o 0} frac{sin x}{x} = 1$.
(The proof extends to $x < 0$ by considering $y = -x$).
* $lim_{x o 0} frac{ an x}{x} = 1$
* $lim_{x o 0} frac{1 - cos x}{x^2} = frac{1}{2}$
(Can be derived from $frac{1-cos x}{x^2} = frac{2sin^2(x/2)}{x^2} = frac{1}{2} left(frac{sin(x/2)}{x/2}
ight)^2 o frac{1}{2}(1)^2 = frac{1}{2}$)
* $lim_{x o 0} frac{sin^{-1} x}{x} = 1$
* $lim_{x o 0} frac{ an^{-1} x}{x} = 1$
C. Exponential and Logarithmic Standard Limits:
* $lim_{x o 0} frac{e^x - 1}{x} = 1$
* $lim_{x o 0} frac{a^x - 1}{x} = ln a$ (where $a > 0, a
e 1$)
* $lim_{x o 0} frac{ln(1+x)}{x} = 1$
*
$lim_{x o 0} (1+x)^{1/x} = e$
*
$lim_{x o infty} (1+frac{1}{x})^x = e$
(These last two are forms of $1^infty$ which defines $e$)
* More generally, if $lim_{x o a} f(x) = 1$ and $lim_{x o a} g(x) = infty$, then $lim_{x o a} [f(x)]^{g(x)} = e^{lim_{x o a} [f(x)-1]g(x)}$. This is extremely useful for $1^infty$ indeterminate forms.
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### 6. The Squeeze Play Theorem (Sandwich Theorem)
We briefly used this theorem in the derivation of $lim_{x o 0} frac{sin x}{x}$. It's a powerful tool for finding limits of functions that are "sandwiched" between two other functions whose limits are known and equal.
Theorem Statement:
If $g(x) le f(x) le h(x)$ for all $x$ in some open interval containing $a$ (except possibly at $a$ itself), and if
$lim_{x o a} g(x) = L$ and $lim_{x o a} h(x) = L$,
then $lim_{x o a} f(x) = L$.
Intuition: Imagine you have two friends, one always walking ahead of you, and one always walking behind you. If both your friends are walking towards the same destination, and you are always between them, then you must also be walking towards that same destination.
Example 5: Using Sandwich Theorem
Find $lim_{x o 0} x sin(frac{1}{x})$.
1. We know that for any real number $y$, $-1 le sin y le 1$.
2. So, for $y = frac{1}{x}$, we have $-1 le sin(frac{1}{x}) le 1$.
3. Now, we need to multiply this inequality by $x$. We must consider two cases:
*
Case 1: $x > 0$ (approaching 0 from the right)
Multiplying by $x$ (a positive number), the inequalities remain unchanged:
$-x le x sin(frac{1}{x}) le x$.
As $x o 0^+$, $lim_{x o 0^+} (-x) = 0$ and $lim_{x o 0^+} x = 0$.
By Sandwich Theorem, $lim_{x o 0^+} x sin(frac{1}{x}) = 0$.
*
Case 2: $x < 0$ (approaching 0 from the left)
Multiplying by $x$ (a negative number), the inequalities reverse:
$-x ge x sin(frac{1}{x}) ge x$.
This can be rewritten as $x le x sin(frac{1}{x}) le -x$.
As $x o 0^-$, $lim_{x o 0^-} x = 0$ and $lim_{x o 0^-} (-x) = 0$.
By Sandwich Theorem, $lim_{x o 0^-} x sin(frac{1}{x}) = 0$.
4. Since LHL = RHL = 0, the limit $lim_{x o 0} x sin(frac{1}{x}) = 0$.
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### 7. Limits Involving Greatest Integer Function (GIF) and Fractional Part Function (FPF)
These functions often cause limits to not exist because they introduce jumps or discontinuities. You MUST evaluate LHL and RHL separately.
*
Greatest Integer Function ($[x]$): The largest integer less than or equal to $x$.
* $lim_{x o n^-} [x] = n-1$ (where $n$ is an integer)
(e.g., $lim_{x o 3^-} [x] = [2.999...] = 2$)
* $lim_{x o n^+} [x] = n$ (where $n$ is an integer)
(e.g., $lim_{x o 3^+} [x] = [3.000...] = 3$)
Since LHL $
e$ RHL at integer points, $lim_{x o n} [x]$ does not exist for integer $n$.
*
Fractional Part Function (${x}$): ${x} = x - [x]$.
* $lim_{x o n^-} {x} = lim_{x o n^-} (x - [x]) = n - (n-1) = 1$
(e.g., $lim_{x o 3^-} {x} = 3 - [2.999...] = 3 - 2 = 1$)
* $lim_{x o n^+} {x} = lim_{x o n^+} (x - [x]) = n - n = 0$
(e.g., $lim_{x o 3^+} {x} = 3 - [3.000...] = 3 - 3 = 0$)
Since LHL $
e$ RHL at integer points, $lim_{x o n} {x}$ does not exist for integer $n$.
Example 6: Evaluate $lim_{x o 1} frac{[x]-1}{x-1}$.
1.
LHL at $x=1$:
$lim_{x o 1^-} frac{[x]-1}{x-1}$
As $x o 1^-$, $x$ is slightly less than 1, so $[x] = 0$.
$= lim_{x o 1^-} frac{0-1}{x-1} = lim_{x o 1^-} frac{-1}{x-1}$
As $x o 1^-$, $x-1$ is a small negative number (e.g., -0.0001).
So, $frac{-1}{ ext{small negative number}} = infty$.
2.
RHL at $x=1$:
$lim_{x o 1^+} frac{[x]-1}{x-1}$
As $x o 1^+$, $x$ is slightly greater than 1, so $[x] = 1$.
$= lim_{x o 1^+} frac{1-1}{x-1} = lim_{x o 1^+} frac{0}{x-1} = 0$.
Since LHL ($infty$) $
e$ RHL (0), the limit $lim_{x o 1} frac{[x]-1}{x-1}$
does not exist.
---
### 8. Limits at Infinity
These limits deal with the behavior of a function as $x$ becomes extremely large (positive or negative). We write these as $lim_{x o infty} f(x)$ or $lim_{x o -infty} f(x)$.
Key Idea: For rational functions (polynomials divided by polynomials), divide both the numerator and denominator by the highest power of $x$ in the denominator.
Example 7: Limits at Infinity
Evaluate $lim_{x o infty} frac{3x^2 + 2x - 1}{5x^2 - 4x + 7}$.
1. Identify the highest power of $x$ in the denominator, which is $x^2$.
2. Divide every term in the numerator and denominator by $x^2$:
$lim_{x o infty} frac{frac{3x^2}{x^2} + frac{2x}{x^2} - frac{1}{x^2}}{frac{5x^2}{x^2} - frac{4x}{x^2} + frac{7}{x^2}}$
$= lim_{x o infty} frac{3 + frac{2}{x} - frac{1}{x^2}}{5 - frac{4}{x} + frac{7}{x^2}}$
3. As $x o infty$, terms like $frac{c}{x^n}$ (where $n > 0$) approach 0.
So, $lim_{x o infty} frac{2}{x} = 0$, $lim_{x o infty} frac{1}{x^2} = 0$, $lim_{x o infty} frac{4}{x} = 0$, $lim_{x o infty} frac{7}{x^2} = 0$.
4. Therefore, the limit becomes:
$frac{3 + 0 - 0}{5 - 0 + 0} = frac{3}{5}$.
CBSE vs JEE Focus: While CBSE covers basic limit evaluation, the $epsilon$-$delta$ definition, the Squeeze Theorem with complex functions, and intricate limits involving GIF/FPF, along with advanced $1^infty$ forms, are more aligned with JEE Advanced preparation. Mastering these deep concepts gives you a significant edge.
This detailed exploration should equip you with a strong conceptual foundation and the necessary tools to tackle a wide range of limit problems, from straightforward evaluations to complex JEE-level challenges. Keep practicing, and remember the core ideas: "approaching," "tendency," and the strictness of the $epsilon$-$delta$ definition.