Factorisation

This is one of the most useful "standard limits" because it gives you a shortcut to solve problems that would otherwise require long and difficult factorization.

1. The Formula

The theorem states: $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$ Where:

• x –> a: The variable x approaches a constant value a.
• n: Is a power (exponent). The image notes that while the proof is for positive integers, the formula actually works even if n is a fraction (rational number).

2. Why do we need this?

If you try to solve this limit by Direct Substitution (plugging in x = a), you get: $\frac{a^n - a^n}{a - a} = \frac{0}{0}$

This is an Indeterminate Form. Usually, you would have to factorize the numerator to cancel out the (x-a) term. This theorem does that work for you instantly.

3. Explanation of the Proof (From the Image)

The proof uses an algebraic identity to factorize $x^n - a^n.$

Step 1: The Expansion

You might remember that $x^2 - a^2 = (x-a)(x+a)$ Or that $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$ There is a general pattern for any power n: $x^n - a^n = (x-a) (x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + a^{n-1})$

Notice the pattern in the second bracket: The power of x goes down by 1 each time, and the power of a goes up by 1.

Step 2: Cancellation

When we divide by (x-a), the term causing the zero cancels out: $\frac{x^n - a^n}{x - a} = \underbrace{x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + a^{n-1}}_{\text{The remaining long polynomial}}$

Step 3: Applying the Limit

Now, we can safely plug in x = a into the remaining polynomial. Every term becomes $a^{n-1}$:

• First term: $a^{n-1}$
• Second term: $a^{n-2} \cdot a^1 = a^{n-1}$
• Third term: $a^{n-3} \cdot a^2 = a^{n-1}$
• Last term: $a^{n-1}$

Step 4: The Final Count

Since the polynomial started with degree n-1 and went down to 0, there are exactly n terms in this series. So, you have $a^{n-1}$ added to itself n times.

Result

${Result} = n \cdot a^{n-1}$

Indeterminate forms

• Think of an "Indeterminate Form" like a mystery box in mathematics.
• In plain English, "Indeterminate" means "cannot be determined yet."
• It doesn't mean the answer doesn't exist; it just means the current form of the expression isn't giving you enough information to find the answer. You have to "open the box" (do more math) to see what's inside. Here is the simplest way to understand the most common indeterminate form, $\frac{0}{0}$

1. The Conflict of Rules

In math, we have three very strong rules for division. When you see $\frac{0}{0}$, these rules start fighting each other:

• Rule A says: "Zero divided by anything is 0" (e.g.,$\frac{0}{5} = 0$).
• Rule B says: "Anything divided by zero is undefined (or huge)" (e.g., $\frac{5}{0}$ is undefined).
• Rule C says: "Anything divided by itself is 1" (e.g., $\frac{5}{5} = 1$). So, when you have $\frac{0}{0}$, which rule wins? Is the answer 0? Is it undefined? Or is it 1? Answer: It could be any of them. We don't know yet. That is why it is "Indeterminate."

2. See it in Action (The best proof)

Let's look at three different limit problems. All of them look like $\frac{0}{0}$ if you just plug in the number, but they have completely different answers. Imagine x is getting extremely close to 0.

Case 1: The answer is 1

$\lim_{x \to 0} \frac{x}{x}$

• Plug in 0: You get $\frac{0}{0}.$
• But simplify it: $\frac{x}{x}$ is just 1.
• Limit = 1

Case 2: The answer is 2

$\lim_{x \to 0} \frac{2x}{x}$

• Plug in 0: You get $\frac{2(0)}{0}$ which is $\frac{0}{0}$
• But simplify it: The x's cancel out, leaving just 2.
• Limit = 2

Case 3: The answer is 0

$\lim_{x \to 0} \frac{x^2}{x}$

• Plug in 0: You get $\frac{0}{0}$
• But simplify it: $\frac{0}{0}$

• If x goes to 0, the answer is 0.

• Limit = 0

Summary

In all three cases, the "form" was $\frac{0}{0}$ but the actual results were 1, 2, and 0. So, whenever you are solving a limit problem and you get $\frac{0}{0}$, treat it as a Red Stop Sign. It is telling you:

"Stop! Don't write zero or undefined. The answer is hidden. You need to factorize, simplify, or use a formula (like the one we just discussed) to find the real value."

Exercise 12.1

Find $\lim_{x \to \pi} (x - \frac{22}{7})$

$\lim_{x \to \pi} \left(x - \frac{22}{7}\right) = \pi - \frac{22}{7}$

Important Note: You might be tempted to say the answer is 0 because often in lower grades we use $\pi \approx \frac{22}{7}$. However, in calculus and higher math, $\pi$ is an irrational number and is not exactly equal to $\frac{22}{7}$. Therefore, the most accurate answer is simply ($\pi - \frac{22}{7}).$

Question 13:

Evaluate the limit: $\lim_{x \to 0} \frac{\sin(ax)}{bx}$

Solution:

Step 1: Identify the Standard Limit To solve this, we use the fundamental trigonometric limit theorem: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ Note: For this theorem to work, the term in the denominator must be exactly the same as the angle inside the sine function. Step 2: Analyze our Problem In our problem, the angle is $ax$, but the denominator is $bx$. $\frac{\sin(ax)}{bx}$ We need the denominator to be $ax$ to match the angle. Step 3: Adjust the Expression We can rewrite the fraction to separate the constant $b$ and introduce the constant $a$ where we need it.

1. Factor out the $\frac{1}{b}$ since it is just a constant: $\lim_{x \to 0} \frac{1}{b} \left( \frac{\sin(ax)}{x} \right)$
2. Multiply the numerator and the denominator by $a$ so that we get $ax$ in the bottom: $\lim_{x \to 0} \frac{1}{b} \cdot \frac{\sin(ax)}{x} \cdot \frac{a}{a}$
3. Rearrange the terms to group $\frac{\sin(ax)}{ax}$ together: $\lim_{x \to 0} \frac{a}{b} \left( \frac{\sin(ax)}{ax} \right)$ Step 4: Apply the Limit Now, as $x \to 0$, naturally $ax \to 0$. According to the standard limit theorem, the term inside the bracket becomes 1. $= \frac{a}{b} \times 1$

Final Answer:

$\frac{a}{b}$ (Note: This solution assumes $b \neq 0$.)

Types of questions in 12.1

Type 1: Direct Substitution

These are the simplest problems. You just plug the value of $x$ into the function. If you get a valid number (not $\frac{0}{0}$), that is your answer.

• Questions: 1, 2, 3, 4, 5, 9, 11
• Special Mention: Question 16 ($\lim_{x \to 0} \frac{\cos x}{\pi - x}$) and Question 19 ($\lim_{x \to 0} x \sec x$) are also solved by direct substitution, even though they look like trigonometric limit problems.

Type 2: Factorization & Simplification (\frac{0}{0} Form)

When you plug in $x$, you get $\frac{0}{0}$. You must factorize the numerator and denominator to cancel out the common term (usually $x-a$).

• Questions: 7, 8, 12
• Method: Factorize quadratics or simplify complex fractions (like in Q12) until the zero-causing term is gone.

Type 3: Standard Algebraic Limit Theorem

These questions are solved using the formula $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$.

• Questions: 6, 10
• Note: In Q6 and Q10, you often need to use a substitution (e.g., let $x+1 = y$) to make the question look like the formula.

Type 4: Trigonometric Limits

These require the Sandwich Theorem results: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$.

• Questions: 13, 14, 15, 17, 18, 20, 21, 22
• Key Strategy:
  • Q13, 14, 18, 20: Divide numerator and denominator by $x$ or adjust constants ($ax$, $bx$).
  • Q17, 21: Use trigonometric identities (like $1 - \cos 2x = 2\sin^2 x$) to simplify first.
  • Q22: Requires substitution (let $x - \frac{\pi}{2} = h$) to change the limit to $h \to 0$.

Type 5: Left Hand Limit (LHL) & Right Hand Limit (RHL)

These questions involve Piecewise Functions (defined differently for x < a and $x > a$) or Modulus Functions ($|x|$). You must check if LHL = RHL

• Questions: 23, 24, 25, 26, 27, 28, 29, 30, 31, 32
• Focus:
  • Q23–24: Simple piecewise functions.
  • Q25–27: Modulus functions (remember $|x| = -x$ when x < 0).
  • Q28, 32: Finding unknown constants ($a, b, m, n$) given that the limit exists

12.2

Type 1: Derivative from First Principles (The "Long Method")

These are the most critical questions for school exams. The question will explicitly say "from first principles." You cannot use the short-cut formulas here; you must use the limit definition:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

• Questions: 4, 10
  • Q4: Covers algebraic functions ($x^3 - 27$, $\frac{1}{x^2}$, etc.).
  • Q10: Covers trigonometric functions ($\cos x$). This is a very common exam question.
  • Strategy: Be careful with your algebra when expanding $(x+h)^n$ or using trigonometric identities like $\cos C - \cos D$.

Type 2: Derivative at a Specific Point

These questions ask for the value of the derivative at a specific number (e.g., at $x = 10$). You first find the general derivative $f'(x)$ and then plug in the value.

• Questions: 1, 2, 3
  • Q1: Derivative of $x^2 - 2$ at $x=10$.
  • Q2: Derivative of $99x$ at $x=100$.
  • Q3: Derivative of $x$ at $x=1$.
  • Strategy: These are usually 1-mark questions meant to check if you understand that the derivative is a function itself.

Type 3: Algebra of Derivatives (Product & Quotient Rules)

These questions test your ability to handle complex functions where terms are multiplied or divided. You need to use the Product Rule $(uv)' = u'v + uv'$ and the Quotient Rule $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$.

• Questions: 7, 8, 9
  • Q7: Polynomials that need expansion or Product Rule (e.g., $(x-a)(x-b)$).
  • Q8: A rational function $\frac{x^n - a^n}{x-a}$ (Quotient Rule).
  • Q9: A mix of simple power rule and quotient rule problems.
  • Strategy: For Q7, it is often easier to expand the brackets first and then differentiate, rather than using the Product Rule.

Type 4: Trigonometric Derivatives & Special Polynomials

These questions focus on standard results for trig functions ($\sin x, \cos x, \tan x$, etc.) or specific polynomial patterns.

• Questions: 5, 6, 11
  • Q5: A specific proof involving a long polynomial sum.
  • Q6: General power rule application ($x^n + ax^{n-1} \dots$).
  • Q11: Derivatives of trigonometric functions (e.g., $\sin x \cos x$, $5 \sec x + 4 \cos x$).
  • Strategy: Memorize the standard derivatives:
    • $(\sin x)' = \cos x$
    • $(\cos x)' = -\sin x$
    • $(\tan x)' = \sec^2 x$

Summary Table

Answers for 12.1

Here are the step-by-step solutions for Questions 23 to 32 from Exercise 12.1 (formerly 13.1) of NCERT Class 11 Mathematics. These questions focus on Left Hand Limits (LHL) and Right Hand Limits (RHL).

• Rule: For a limit $\lim_{x \to a} f(x)$ to exist, you must show that $\text{LHL} = \text{RHL}$.

Question 23

Find $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$, where $f(x) = \begin{cases} 2x + 3 & x \le 0 \\ 3(x+1) & x > 0 \end{cases}$

Solution: 1. For $\lim_{x \to 0} f(x)$:

Since the definition splits at $x=0$, we must check LHL and RHL.

• LHL (at $x \to 0^-$): Use $f(x) = 2x + 3$.

  $\lim_{x \to 0^-} (2x + 3) = 2(0) + 3 = 3$

• RHL (at $x \to 0^+$): Use $f(x) = 3(x+1)$.

  $\lim_{x \to 0^+} 3(x+1) = 3(0+1) = 3(1) = 3$

• Conclusion: Since $\text{LHL} = \text{RHL} = 3$, the limit exists.

  $\lim_{x \to 0} f(x) = \mathbf{3}$

2. For $\lim_{x \to 1} f(x)$:

The point $x=1$ is strictly greater than 0. The function definition doesn't change around 1 (it is always $3(x+1)$ for $x>0$).

• $\lim_{x \to 1} f(x) = \lim_{x \to 1} 3(x+1) = 3(1+1) = 3(2) = \mathbf{6}$

Question 24

Find $\lim_{x \to 1} f(x)$, where

$f(x) = \begin{cases} x^2 - 1 & x \le 1 \\ -x^2 - 1 & x > 1 \end{cases}$

Solution:

Check LHL and RHL at $x=1$.

• LHL (at $x \to 1^-$): Use $f(x) = x^2 - 1$.

  $\lim_{x \to 1^-} (x^2 - 1) = (1)^2 - 1 = 1 - 1 = 0$

• RHL (at $x \to 1^+$): Use $f(x) = -x^2 - 1$.

  $\lim_{x \to 1^+} (-x^2 - 1) = -(1)^2 - 1 = -1 - 1 = -2$

• Conclusion: Since $\text{LHL} (0) \neq \text{RHL} (-2)$, the limit does not exist.

Question 25

Evaluate $\lim_{x \to 0} f(x)$, where

$f(x) = \begin{cases} \frac{|x|}{x} & x \neq 0 \\ 0 & x = 0 \end{cases}$

Solution:

Recall that $|x| = -x$ for x < 0 and $|x| = x$ for $x > 0$. The Definition:

• Case 1 (Positive Zone): If $x > 0$, we leave it alone. So, $|x| = x$.
• Case 2 (Negative Zone): If x < 0, we must multiply by $-1$ to fix the sign. So, $|x| = -x$.

Key Takeaway: The "$-x$" definition for negative numbers doesn't mean the result is negative. It means "the negative of the negative number," which makes it positive.

• LHL (at $x \to 0^-$): Here $|x| = -x$. $\lim_{x \to 0^-} \frac{-x}{x} = \lim_{x \to 0^-} (-1) = -1$

• RHL (at $x \to 0^+$): Here $|x| = x$.

• $\lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} (1) = 1$

• Conclusion: Since $\text{LHL} (-1) \neq \text{RHL} (1)$, the limit does not exist.

Question 26

Find $\lim_{x \to 0} f(x)$, where

$f(x) = \begin{cases} \frac{x}{|x|} & x \neq 0 \\ 0 & x = 0 \end{cases}$

Solution:

This is very similar to Question 25.

• LHL (at $x \to 0^-$): Here $|x| = -x$.

  $\lim_{x \to 0^-} \frac{x}{-x} = -1$

• RHL (at $x \to 0^+$): Here $|x| = x$. $\lim_{x \to 0^+} \frac{x}{x} = 1$

• Conclusion: $\text{LHL} \neq \text{RHL}$, so the limit does not exist.

Question 27

Find $\lim_{x \to 5} f(x)$, where $f(x) = |x| - 5$. Solution: Here, the function definition $|x| - 5$ is continuous and doesn't change its formula around $x=5$ (since 5 is positive, $|x|=x$). At $x \to 5$ $\to$ The neighbors are 4.99 and 5.01. Both are positive. The function behaves normally, so we just plug in the number. $\lim_{x \to 5} (|x| - 5) = |5| - 5 = 5 - 5 = \mathbf{0}$

Question 28

Suppose $f(x) = \begin{cases} a + bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{cases}$. If $\lim_{x \to 1} f(x) = f(1)$, what are possible values of $a$ and $b$?

Solution:

Given condition: $\lim_{x \to 1} f(x) = f(1) = 4$.

This means both LHL and RHL at $x=1$ must equal 4.

1. LHL must be 4:

   $\lim_{x \to 1^-} (a + bx) = a + b(1) = a + b$

   So, $a + b = 4$ — (Equation 1)

2. RHL must be 4:

   $\lim_{x \to 1^+} (b - ax) = b - a(1) = b - a$

   So, $b - a = 4$ — (Equation 2)

3. Solve the system: From Eq 2: $b = 4 + a$. Substitute into Eq 1: $a + (4 + a) = 4$ $2a + 4 = 4 \implies 2a = 0 \implies \mathbf{a = 0}$ Substitute $a=0$ into Eq 1: $0 + b = 4 \implies \mathbf{b = 4}$ Answer: $a = 0, b = 4$.

Question 29

Let $a_1, a_2, ..., a_n$ be fixed real numbers and $f(x) = (x-a_1)(x-a_2)...(x-a_n)$. Find $\lim_{x \to a_1} f(x)$ and $\lim_{x \to a} f(x)$. Solution:

1. For $\lim_{x \to a_1} f(x)$: Substitute $x = a_1$ directly. $= (a_1 - a_1)(a_1 - a_2)...(a_1 - a_n)$ $= 0 \cdot (\text{other terms}) = \mathbf{0}$
2. For $\lim_{x \to a} f(x)$ (where $a \neq a_i$): Simply substitute $x = a$. $= \mathbf{(a - a_1)(a - a_2)...(a - a_n)}$

Question 30

If $f(x) = \begin{cases} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| - 1 & x > 0 \end{cases}$. For what value(s) of $a$ does $\lim_{x \to a} f(x)$ exist? Solution: Let's check the critical point $x=0$ where the definition changes.

• LHL at 0: $\lim_{x \to 0^-} (|x| + 1) = |-0| + 1 = 0 + 1 = 1$.
• RHL at 0: $\lim_{x \to 0^+} (|x| - 1) = |0| - 1 = 0 - 1 = -1$.
• Since $1 \neq -1$, the limit does not exist at $a=0$.

For any other point ($a \neq 0$):

• If a < 0, $f(x) = -x+1$ (polynomial), so limit exists.
• If $a > 0$, $f(x) = x-1$ (polynomial), so limit exists.

Answer: The limit exists for all real numbers $a$ except $a = 0$.

Question 31

If $\lim_{x \to 1} \frac{f(x) - 2}{x^2 - 1} = \pi$, evaluate $\lim_{x \to 1} f(x)$.

Solution:

We are given that the limit of the fraction exists and is equal to $\pi$.

As $x \to 1$, the denominator $(x^2 - 1) \to (1^2 - 1) = 0$.

For the answer to be a finite number ($\pi$) and not infinity, the numerator must also be zero (creating a $\frac{0}{0}$ form).

Therefore:

$\lim_{x \to 1} (f(x) - 2) = 0$

$\lim_{x \to 1} f(x) = \mathbf{2}$

---

Question 32

If $f(x) = \begin{cases} mx^2 + n & x < 0 \\ nx + m & 0 \le x \le 1 \\ nx^3 + m & x > 1 \end{cases}$. For what integers $m$ and $n$ does both $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$ exist?

Solution: 1. Check Limit at $x = 0$:

• LHL (x < 0): $mx^2 + n \to m(0)^2 + n = n$.
• RHL ($x > 0$): $nx + m \to n(0) + m = m$.
• For limit to exist: $\text{LHL} = \text{RHL} \implies \mathbf{n = m}$. 2. Check Limit at $x = 1$:
• LHL (x < 1): $nx + m \to n(1) + m = n + m$.
• RHL ($x > 1$): $nx^3 + m \to n(1)^3 + m = n + m$.
• $\text{LHL} = \text{RHL}$ is always true ($n+m = n+m$). So, the limit at $x=1$ exists for any values of $m$ and $n$.

Conclusion: For the limit to exist at both points, the condition from step 1 must be satisfied. Answer: The limits exist for any integral values of $m$ and $n$ such that $m = n$.

12.2 Derivatives

In Class 11 calculus, the First Principle of Derivative (often called the definition of the derivative) is the formal way we define the rate of change of a function. It is the foundation upon which all other shortcut rules (like the power rule) are built.

1. The Core Concept: Slope of a Tangent

The derivative $f'(x)$ represents the slope of the tangent line to a curve at any point.

Imagine a curve $f(x)$. To find the slope at point $P(x, f(x))$, we take a second point $Q$ very close to it, at $(x+h, f(x+h))$, where $h$ is a tiny distance. The slope of the line connecting them (the secant) is:

$\text{Slope} = \frac{\text{Change in } y}{\text{Change in } x} = \frac{f(x+h) - f(x)}{h}$

As we make $h$ smaller and smaller (approaching zero), the point $Q$ moves closer to $P$, and the secant line becomes the tangent line.

2. The Formula

The official formula for the first principle is: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

3. Step-by-Step Example: $f(x) = x^2$

To find the derivative of $x^2$ using this rule, we follow these steps:

1. Identify $f(x+h)$: Replace $x$ with $(x+h)$. $f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$
2. Plug into the formula: $f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - (x^2)}{h}$
3. Simplify the numerator: The $x^2$ and $-x^2$ cancel out. $f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}$
4. Divide by $h$: $f'(x) = \lim_{h \to 0} (2x + h)$
5. Apply the limit ($h = 0$): $f'(x) = 2x$

The Logic of "Approaching" Zero

When we say $\lim_{h \to 0}$, we aren't saying $h$ is zero yet. We are saying $h$ is becoming so incredibly small ($0.0000001$) that the second point $Q$ is practically on top of the first point $P$.

Once the $h$ in the denominator is cancelled out through simplification, the "threat" of an undefined fraction is gone. Now, we can safely let $h$ become $0$.

Common First Principle Derivatives

Graph

![obsidian|300][Derivatives.webp]

Answers for 12.2

10. Derivative of Cosx by First Principle

$\frac {d}{dx} ​cos(x) = lim_{h→0} \frac {cos⁡(x+h)−cos⁡(x)}{h}​$

Using the angle addition formula for cosine:

= $lim_{h → 0} \frac {(cos⁡(x) \times cos⁡(h) − sin⁡(x) \times sin⁡(h)) − cos⁡(x)}{h}$​

= $lim_{h → 0} \frac {(cos⁡(x)(cos⁡(h)−1)) − sin⁡(x)sin⁡(h)}{h}$​

= $cos(x) lim_{h → 0} \frac {(cos⁡(h)−1)}{h} - sin(x) lim_{h → 0} \frac {sin⁡(h)}{h}$​

As $lim_{h → 0} \frac {cos⁡(h)−1}{h} = 0)$ and $lim_{h → 0} \frac {sin⁡(h)}{h}​ = 1$

= cos(x) ⋅ 0 - sin(x) ⋅ 1

= -sin(x)

• So, the derivative of cos(x) with respect to ( x ) using the first principles of derivatives is -sin(x).