**Chapter 12.1: Introduction**

**Limits:**The concept of a limit is central to calculus and is used to describe the behavior of a function as its input approaches a certain value.**Derivatives:**Derivatives measure the rate of change of a function. They are essential for understanding concepts like velocity, acceleration, and slopes of curves.

**Chapter 12.2: Limits of Functions**

**Definition of Limit:**The limit of a function f(x) as x approaches a is denoted by lim(x→a) f(x). It represents the value that f(x) approaches as x gets closer and closer to a.**Left-hand and Right-hand Limits:**These are limits taken from the left and right sides of a point, respectively.**Existence of Limits:**For a limit to exist, the left-hand and right-hand limits must be equal.**Limit Laws:**These are rules for evaluating limits of algebraic combinations of functions.

**Chapter 12.3: Evaluation of Limits**

**Methods of Evaluation:**Direct substitution, factorization, rationalization, and using standard limits.**Standard Limits:**Important limits such as lim(x→0) sin(x)/x = 1 and lim(x→0) (1 – cos(x))/x = 0.

**Chapter 12.4: Derivative**

**Notation:**f'(a) or dy/dx.**Geometric Interpretation:**The derivative at a point represents the slope of the tangent line to the graph of f(x) at that point.

**Chapter 12.5: Differentiation**

**Rules of Differentiation:**Power rule, sum rule, difference rule, product rule, quotient rule, chain rule.**Differentiation of Trigonometric Functions:**Derivatives of sin(x), cos(x), tan(x), cot(x), sec(x), and csc(x).

**Key Concepts:**

- Limits and their evaluation
- Derivatives and their geometric interpretation
- Rules of differentiation
- Applications of derivatives (e.g., finding velocity, acceleration, tangent lines)

**Exercise 12.1 **

**Evaluate the following limits in Exercises 1 to 22.**

**1. **

** **

**Ans : **

To evaluate the limit lim(x→3) (x + 3), we can simply substitute 3 for x:

lim(x→3) (x + 3) = 3 + 3 = 6

Therefore, the limit of x + 3 as x approaches 3 is 6.

**2. **

**Ans : **

To evaluate the limit, we can directly substitute π for x:

lim(x→π) (x – 22/7)

= π – 22/7

Therefore, the limit of (x – 22/7) as x approaches π is π – 22/7.

**3.**

**Ans : **

To evaluate the limit, we can simply substitute 1 for r:

lim(r→1) πr^2 = π(1)^2 = π

Therefore, the limit of πr^2 as r approaches 1 is π.

**4.**

**Ans : **To evaluate the limit, we can directly substitute 4 for x:

lim(x→4) (4x + 3) / (x – 2) = (4 * 4 + 3) / (4 – 2) = (16 + 3) / 2 = 19 / 2

Therefore, the limit of (4x + 3) / (x – 2) as x approaches 4 is 19/2

**5. **

**Ans : **To evaluate the limit, we can’t directly substitute -1 for x because it would result in an indeterminate form of 0/0. Instead, we can try factoring the numerator.

Notice that x^10 + x^5 + 1 can be factored as (x^5 + 1)^2.

So, the limit becomes:

lim(x→-1) [(x^5 + 1)^2 / (x – 1)]

Now, we can use direct substitution:

lim(x→-1) [(x^5 + 1)^2 / (x – 1)] = [(-1)^5 + 1]^2 / (-1 – 1) = 0 / (-2) = 0

Therefore, the limit of (x^10 + x^5 + 1) / (x – 1) as x approaches -1 is 0.

**6.**

**Ans : **

**7.**

**Ans :**

**8.**

**Ans : **

**9.**

**Ans : **

lim(x→0) (ax + b) / (cx + 1) = (a(0) + b) / (c(0) + 1) = b / 1 = b

Therefore, the limit of (ax + b) / (cx + 1) as x approaches 0 is b.

**10.**

**Ans : **

**11.**

**Ans : **

**12.**

**Ans : **

**13.**

**Ans : **

**14.**

**Ans : **

To evaluate the limit, we can use the following trigonometric identity:

lim(x→0) sin(x)/x = 1

We can rewrite the given limit as:

lim(x→0) (sin(ax) / x) / (sin(bx) / x)

Now, using the above identity:

lim(x→0) (sin(ax) / x) / (sin(bx) / x) = (a * 1) / (b * 1) = a/b

Therefore, the limit of sin(ax) / sin(bx) as x approaches 0 is a/b, where a and b are not equal to 0.

**15.**

**Ans : **

**16.**

**Ans : **

**17.**

**Ans : **

**18.**

**Ans : **

To evaluate the limit, we can use the following trigonometric identity:

lim(x→0) sin(x)/x = 1

We can rewrite the given limit as:

lim(x→0) (ax + cos(x)) / (b sin(x))

Now, divide both the numerator and denominator by x:

= lim(x→0) ((ax/x) + (cos(x)/x)) / (b sin(x)/x)

Using the trigonometric identity:

= lim(x→0) (a + cos(x)/x) / (b * 1)

As x approaches 0, cos(x)/x approaches 1. Therefore, the limit becomes:

= (a + 1) / b

**So, the limit of (ax + cos(x)) / (b sin(x)) as x approaches 0 is (a + 1) / b.**

**19.**

**Ans :**

**20**.

**Ans : **

To evaluate the limit, we can use the following trigonometric identity:

lim(x→0) sin(x)/x = 1

We can rewrite the given limit as:

lim(x→0) (sin(ax) + bx) / (ax + sin(bx))

Now, divide both the numerator and denominator by x:

= lim(x→0) ((sin(ax)/x) + (bx/x)) / ((ax/x) + (sin(bx)/x))

Using the trigonometric identity:

= lim(x→0) ((a * 1) + b) / ((a * 1) + (b * 1))

Simplifying:

= (a + b) / (a + b)

= 1

**21.**

**Ans : **

**22.**

**Ans :**

**23.**

**Ans : **

**24.**

**Ans : **

**For lim(x→1-) f(x):**

Since x is approaching 1 from the left, we use the definition of f(x) for x ≤ 1:

lim(x→1-) f(x) = lim(x→1-) (x^2 – 1)

Direct substitution gives us:

= 1^2 – 1 = 0

**For lim(x→1+) f(x):**

Since x is approaching 1 from the right, we use the definition of f(x) for x > 1:

lim(x→1+) f(x) = lim(x→1+) (-x^2 – 1)

Direct substitution gives us:

= -(1^2) – 1 = -2

**25.**

**Ans : **

**For lim(x→0-) f(x):**

Since x is approaching 0 from the left, we use the definition of f(x) for x ≠ 0:

lim(x→0-) f(x) = lim(x→0-) |x|/x

For x < 0, |x| = -x. Substituting this:

= lim(x→0-) -x/x = lim(x→0-) -1 = -1

**For lim(x→0+) f(x):**

Since x is approaching 0 from the right, we use the definition of f(x) for x ≠ 0:

lim(x→0+) f(x) = lim(x→0+) |x|/x

For x > 0, |x| = x. Substituting this:

= lim(x→0+) x/x = lim(x→0+) 1 = 1

**26.**

**Ans : **

**For lim(x→0-) f(x):**

Since x is approaching 0 from the left, we use the definition of f(x) for x ≠ 0:

lim(x→0-) f(x) = lim(x→0-) |x|/x

For x < 0, |x| = -x. Substituting this:

= lim(x→0-) -x/x = lim(x→0-) -1 = -1

**For lim(x→0+) f(x):**

Since x is approaching 0 from the right, we use the definition of f(x) for x ≠ 0:

lim(x→0+) f(x) = lim(x→0+) |x|/x

For x > 0, |x| = x. Substituting this:

= lim(x→0+) x/x = lim(x→0+) 1 = 1

**27.**

**Ans : **

**28.**

**Ans : **

**Given:**

- f(x) = { a + bx, x < 1 { 4, x = 1 { b – ax, x > 1
- lim(x→1) f(x) = f(1) = 4

**Solution:**

**1. Evaluate the limits from both sides:**

**Left-hand limit:**lim(x→1-) f(x) = lim(x→1-) (a + bx) = a + b**Right-hand limit:**lim(x→1+) f(x) = lim(x→1+) (b – ax) = b – a

**2. Apply the condition for the limit to exist:**

For the limit to exist, the left-hand limit and the right-hand limit must be equal to the function’s value at x = 1. Therefore:

a + b = 4 (from the left-hand limit) b – a = 4 (from the right-hand limit)

**3. Solve the system of equations:**

Adding the two equations:

2b = 8 b = 4

Substituting b = 4 into the first equation:

a + 4 = 4

a = 0

**29.**

**Ans : **

**30.**

**Ans : **

**Case 1: a < 0**

In this case, as x approaches a, it will also be less than 0. Therefore, we use the definition of f(x) for x < 0:

lim(x→a-) f(x) = lim(x→a-) (|x| + 1)

Since x is approaching a from the left, |x| = -x:

= lim(x→a-) (-x + 1)

Substituting x = a:

= -a + 1

**Case 2: a = 0**

In this case, the limit is directly the value of f(0):

lim(x→0) f(x) = f(0) = 0

**Case 3: a > 0**

In this case, as x approaches a, it will also be greater than 0. Therefore, we use the definition of f(x) for x > 0:

lim(x→a+) f(x) = lim(x→a+) (|x| – 1)

= lim(x→a+) (x – 1)

Substituting x = a:

= a – 1

**For the limit to exist, the left-hand limit, right-hand limit, and the function’s value at a must all be equal.**

-a + 1 = 0 = a – 1

Solving the first equation:

-a + 1 = 0 a = 1

Solving the second equation:

a – 1 = 0 a = 1

**Therefore, the limit lim(x→a) f(x) exists only when a = 1.**

**31.**

**Ans : **

To evaluate the limit lim(x→1) f(x), we can use the given information about the limit of the quotient (f(x) – 2) / (x^2 – 1).

We know that:

lim(x→1) (f(x) – 2) / (x^2 – 1) = π

We can rewrite the expression (x^2 – 1) as (x – 1)(x + 1). Then, the limit becomes:

lim(x→1) (f(x) – 2) / ((x – 1)(x + 1)) = π

Now, we can use the property of limits that states:

lim(x→a) [f(x) / g(x)]

= [lim(x→a) f(x)] / [lim(x→a) g(x)]

Applying this property to our limit:

[lim(x→1) (f(x) – 2)] / [lim(x→1) (x – 1)(x + 1)] = π

Since lim(x→1) (x – 1)(x + 1) = 0, we can multiply both sides of the equation by this expression:

lim(x→1) (f(x) – 2)

= π * 0

lim(x→1) (f(x) – 2) = 0

lim(x→1) f(x) = 2

Therefore, the limit of f(x) as x approaches 1 is **2**.

**32.**

**Ans :**

**Exercise 12.2**

**1. Find the derivative of ****x****2****– 2 at x = 10.**

**Ans : **

To find the derivative of x^2 – 2 at x = 10, we’ll first differentiate the function and then evaluate the derivative at x = 10.

**1. Differentiate the function:**

f'(x) = d/dx (x^2 – 2) = 2x – 0 = 2x

**2. Evaluate the derivative at x = 10:**

f'(10) = 2 * 10 = 20

Therefore, the derivative of x^2 – 2 at x = 10 is **20**.

**2. Find the derivative of x at x = 1**

**Ans : **

**The derivative of x at any point is 1.**

This is a fundamental rule of calculus. The derivative of a linear function (like x) is its slope, and the slope of the line y = x is 1.

Therefore, the derivative of x at x = 1 is also **1**.

**3. Find the derivative of 99x at x = l00**

**Ans : **

To find the derivative of 99x at x = 100, we first need to find the general derivative of the function.

The derivative of 99x is 99. This is because the derivative of any linear function ax is a, where a is a constant.

Therefore, the derivative of 99x at x = 100 is also **99**

**4.**

**Ans : **

**(i) f(x) = x^2 – 27**

f'(x) = lim(h→0) [(x + h)^2 – 27 – (x^2 – 27)] / h

= lim(h→0) (x^2 + 2hx + h^2 – 27 – x^2 + 27) / h

= lim(h→0) (2hx + h^2) / h

= lim(h→0) (2x + h)

= 2x

Therefore, the derivative of x^2 – 27 is 2x.

**(ii) f(x) = (x – 1)(x – 2)**

f'(x) = lim(h→0) [(x + h – 1)(x + h – 2) – (x – 1)(x – 2)] / h

= lim(h→0) (x^2 + 2hx + h^2 – 3x + 2 – x^2 + 3x – 2) / h

= lim(h→0) (2hx + h^2) / h

= lim(h→0) (2x + h)

= 2x

Therefore, the derivative of (x – 1)(x – 2) is 2x.

**(iii) f(x) = 1/x^2**

f'(x) = lim(h→0) [1/(x + h)^2 – 1/x^2] / h

= lim(h→0) [(x^2 – (x + h)^2) / (x^2(x + h)^2)] / h

= lim(h→0) (-2hx – h^2) / (x^2(x + h)^2 * h)

= lim(h→0) (-2x – h) / (x^2(x + h)^2)

= -2x / x^4

= -2/x^3

Therefore, the derivative of 1/x^2 is -2/x^3.

**(iv) f(x) = (x + 1)/(x – 1)**

f'(x) = lim(h→0) [(x + h + 1)/(x + h – 1) – (x + 1)/(x – 1)] / h

= lim(h→0) [(x + h + 1)(x – 1) – (x + 1)(x + h – 1)] / ((x – 1)(x + h – 1)h)

= lim(h→0) (-2h) / ((x – 1)(x + h – 1)h)

= lim(h→0) -2 / ((x – 1)(x + h – 1))

= -2 / (x – 1)^2

Therefore, the derivative of (x + 1)/(x – 1) is -2 / (x – 1)^2.

**5.**

**Ans : **

**Given function:**

f(x) = x^100/100 + x^99/99 + … + x^2/2 + x + 1

**To prove:**

f'(1) = 100f'(0)

**Solution:**

**1. Find the derivative of f(x):**

f'(x) = x^99 + x^98 + … + x + 1

**2. Evaluate f'(1):**

f'(1) = 1^99 + 1^98 + … + 1 + 1

= 100

**3. Evaluate f'(0):**

f'(0) = 0^99 + 0^98 + … + 0 + 1

= 1

**4. Compare f'(1) and 100f'(0):**

f'(1) = 100

100f'(0) = 100 * 1 = 100

**Therefore, f'(1) = 100f'(0).**

**6.**

**Ans : **

To find the derivative of the given function, we can use the power rule and the sum rule of differentiation.

The power rule states that the derivative of x^n is nx^(n-1), where n is any real number.

The sum rule states that the derivative of the sum of two functions f(x) and g(x) is the sum of their individual derivatives: (f + g)'(x) = f'(x) + g'(x).

d/dx (x^n + ax^(n-1) + a^2x^(n-2) + … + a^(n-1)x + a^n)

= nx^(n-1) + a(n-1)x^(n-2) + a^2(n-2)x^(n-3) + … + a^(n-1) + 0

= nx^(n-1) + a(n-1)x^(n-2) + a^2(n-2)x^(n-3) + … + a^(n-1)

nx^(n-1) + a(n-1)x^(n-2) + a^2(n-2)x^(n-3) + … + a^(n-1)

7.

**Ans : **

**(i) (x – a)(x – b)**

Using the product rule:

(uv)’ = u’v + uv’

Let u = x – a

and v = x – b.

Then:

u’ = 1

v’ = 1

So, the derivative of (x – a)(x – b) is:

(x – a)(x – b)’ = (x – a)(1) + (x – b)(1) = 2x – a – b

**(ii) (ax^2 + b)^2**

Using the chain rule:

(f(g(x)))’ = f'(g(x)) * g'(x)

Let f(u) = u^2

and g(x) = ax^2 + b.

Then:

f'(u) = 2u

g'(x) = 2ax

So, the derivative of (ax^2 + b)^2 is:

(ax^2 + b)^2′ = 2(ax^2 + b) * 2ax = 4ax(ax^2 + b)

**(iii) (x – a) / (x – b)**

Using the quotient rule:

(u/v)’ = (u’v – uv’) / v^2

Let u = x – a

and v = x – b.

Then:

u’ = 1

v’ = 1

So, the derivative of (x – a) / (x – b) is:

((x – a) / (x – b))’ = ((1)(x – b) – (x – a)(1)) / (x – b)^2

= (x – b – x + a) / (x – b)^2

= (a – b) / (x – b)^2

8.

**Ans :**

**9.**

**Ans : **

**(i) 2x^3/4**

Using the power rule and the constant multiple rule:

d/dx (2x^3/4) = 2 * (3/4) * x^(3/4 – 1) = 3/2 * x^(-1/4)

**(ii) (5x^3 + 3x – 1)(x – 1)**

Using the product rule:

(uv)’ = u’v + uv’

Let u = 5x^3 + 3x – 1 and v = x – 1. Then:

u’ = 15x^2 + 3

v’ = 1

So, the derivative is:

(5x^3 + 3x – 1)(x – 1)’ = (15x^2 + 3)(x – 1) + (5x^3 + 3x – 1)(1)

= 15x^3 – 15x^2 + 3x – 3 + 5x^3 + 3x – 1

= 20x^3 – 15x^2 + 6x – 4

**(iii) x^-3 (5 + 3x)**

Using the product rule:

(uv)’ = u’v + uv’

Let u = x^-3 and v = 5 + 3x. Then:

u’ = -3x^-4

v’ = 3

So, the derivative is:

x^-3 (5 + 3x)’ = (-3x^-4)(5 + 3x) + (x^-3)(3)

= -15x^-4 – 9x^-3 + 3x^-3

= -15x^-4 – 6x^-3

**(iv) x^5 (3 – 6x^9)**

Using the product rule:

(uv)’ = u’v + uv’

Let u = x^5 and v = 3 – 6x^9. Then:

u’ = 5x^4

v’ = -54x^8

So, the derivative is:

x^5 (3 – 6x^9)’ = (5x^4)(3 – 6x^9) + (x^5)(-54x^8)

= 15x^4 – 30x^13 – 54x^13

= 15x^4 – 84x^13

**(v) x^-4 (3 – 4x^5)**

Using the product rule:

(uv)’ = u’v + uv’

Let u = x^-4 and v = 3 – 4x^5. Then:

u’ = -4x^-5

v’ = -20x^4

So, the derivative is:

x^-4 (3 – 4x^5)’ = (-4x^-5)(3 – 4x^5) + (x^-4)(-20x^4)

= 16x^-1 – 12x^-5 – 20x^-1

= -4x^-1 – 12x^-5

**(vi) 2x^2 / (x + 1)(3x – 1)**

Using the quotient rule:

(u/v)’ = (u’v – uv’) / v^2

Let u = 2x^2 and v = (x + 1)(3x – 1). Then:

u’ = 4x

v’ = (1)(3x – 1) + (x + 1)(3) = 6x + 2

So, the derivative is:

(2x^2 / (x + 1)(3x – 1))’ = ((4x)(x + 1)(3x – 1) – (2x^2)(6x + 2)) / ((x + 1)(3x – 1))^2

= (12x^3 – 4x^2 – 4x) / ((x + 1)(3x – 1))^2

**10. Find the derivative of cos x from first principle.**

**Ans : **

To find the derivative of cos x from first principles, we need to use the definition of the derivative:

f'(x) = lim(h→0) [f(x + h) – f(x)] / h

For f(x) = cos x, we have:

f'(x) = lim(h→0) [cos(x + h) – cos(x)] / h

Now, we can use the trigonometric identity:

cos(a + b) = cos(a)cos(b) – sin(a)sin(b)

**To rewrite the expression:**

f'(x) = lim(h→0)

[cos(x)cos(h) – sin(x)sin(h) – cos(x)] / h

**Simplifying:**

f'(x) = lim(h→0) [cos(x)(cos(h) – 1) – sin(x)sin(h)] / h

Now, we can use the following limits:

- lim(h→0) (cos(h) – 1) / h = 0
- lim(h→0) sin(h) / h = 1

**Substituting these limits:**

f'(x) = cos(x) * 0 – sin(x) * 1

Therefore, the derivative of cos x is:

f'(x) = -sin(x)

**11.**

**Ans : **

**(i) sin x cos x**

Using the product rule:

(uv)’ = u’v + uv’

Let u = sin x and v = cos x. Then:

u’ = cos x

v’ = -sin x

So, the derivative is:

(sin x cos x)’ = (cos x)(cos x) + (sin x)(-sin x)

= cos^2(x) – sin^2(x)

= cos(2x)

**(ii) sec x**

Recall that sec x = 1/cos x. Using the quotient rule:

(u/v)’ = (u’v – uv’) / v^2

Let u = 1 and v = cos x. Then:

u’ = 0

v’ = -sin x

So, the derivative is:

(sec x)’ = (0 * cos x – 1 * (-sin x)) / cos^2(x)

= sin x / cos^2(x)

= sec x tan x

**(iii) 5 sec x + 4 cos x**

Using the sum rule and the results from (ii):

(5 sec x + 4 cos x)’ = 5(sec x)’ + 4(cos x)’

= 5(sec x tan x) – 4sin x

= 5sec x tan x – 4sin x

**(iv) cosec x**

Recall that cosec x = 1/sin x. Using the quotient rule:

(u/v)’ = (u’v – uv’) / v^2

Let u = 1 and v = sin x. Then:

u’ = 0

v’ = cos x

So, the derivative is:

(cosec x)’ = (0 * sin x – 1 * cos x) / sin^2(x)

= -cos x / sin^2(x)

= -cosec x cot x

**(v) 3cot x + 5cosec x**

Using the sum rule and the results from (iii) and (iv):

(3cot x + 5cosec x)’ = 3(cot x)’ + 5(cosec x)’

= 3(-csc^2 x) + 5(-cosec x cot x)

= -3csc^2 x – 5cosec x cot x

**(vi) 5sin x – 6cos x + 7**

Using the sum and difference rules and the derivatives of sin x and cos x:

(5sin x – 6cos x + 7)’ = 5(sin x)’ – 6(cos x)’ + 0

= 5cos x + 6sin x

**(vii) 2tan x – 7sec x**

Using the sum rule and the derivatives of tan x and sec x:

(2tan x – 7sec x)’ = 2(tan x)’ – 7(sec x)’

= 2sec^2 x – 7sec x tan x