CBSE Class 11 Maths Chapter 13 - Limits and Derivatives Formulas

Introduction of Limits and Derivatives

Differentiation and calculus fundamentals serve as the basis for advanced mathematics, modern physics and many other modern science and engineering branches. For CBSE students, Class 11 Limits and Derivatives function as the entry point for calculus.

Limits of a Function

In Mathematics, a limit is defined as a value approached as the input by a function, and it produces some value. In calculus and mathematical analysis, limits are important and are used to define integrals, derivatives, and continuity.

Limits Formula

To express a function's limit, we represent it as:

$\lim_{x\to a}f(x)$

Left Hand and Right-Hand Limits

If the function values at the point very close to a, on the left tend to a definite unique number as $x$ tends to $a$, then the unique number so obtained is called the $f(x)$ left-hand limit at $x = a$, we write it as $x = a$.

$f(a-0) = \lim_{x\to a^{-}}f(x) = \lim_{h\to 0}f(a-h)$

Similarly, right hand limit is

$f(a+0) = \lim_{x\to a{+}}f(x) = \lim_{h\to 0}f(a+h)$

Existence of Limit

$\lim_{x\to a}f(x)$ exists, if

(i) $\lim_{x\to a^{-}}f(x)$ and $\lim_{x\to a^{+}}f(x)$ both exists

(ii) $\lim_{x\to a^{-}}f(x) = \lim_{x\to a^{+}}f(x)$

Properties of Limits

1. $lim_{x\to a}[p(x) + g(x)] = \lim_{x\to a}p(x) + \lim_{x\to a}g(x)$

2. $lim_{x\to a}[p(x) - g(x)] = \lim_{x\to a}p(x) - \lim_{x\to a}g(x)$

3. For every real number $k$,

$\lim_{x\to a}[kp(x)] = k \lim_{x\to a}p(x)$

4. $\lim_{x\to a}[p(x) q(x)] = \lim_{x\to a}p(x) \times \lim_{x\to a}q(x)$

5. $\lim_{x\to a}\dfrac{p(x)}{q(x)} = \dfrac{\lim_{x\to a}p(x)}{\lim_{x\to a}q(x)}$

Let two functions be $p$ and $q$ and $a$ value be such that

Derivatives of a Function

A derivative corresponds, in comparison to the other, to the instantaneous rate of change of a quantity. This helps to explore the moment by moment essence of a quantity. A function's derivative is expressed in the formula given below.

Derivative Formula

Assuming f is a real-valued function, then

$f^\prime (x) = \lim_{x\to a}\dfrac{f(x+h)-f(x)}{h}$ is called the derivative of $f$ at $x$ iff $\lim_{x\to a} \dfrac{f(x+h)-f(x)}{h}$ exists finitely.

Its derivative is said to be $f^\prime(x)$ for function $f$, provided that the above equation exists. Here, search all the derivative formulas relating to trigonometric functions, inverse functions, hyperbolic functions, etc.

Properties of Derivatives

1. $\dfrac{d}{dx}[p(x) + q(x)] = \dfrac{d}{dx}(p(x)) + \dfrac{d}{dx}(q(x))$

2. $\dfrac{d}{dx}[p(x) - q(x)] = \dfrac{d}{dx}(p(x)) - \dfrac{d}{dx}(q(x))$

3. $\dfrac{d}{dx}[p(x) \times q(x)] = \dfrac{d}{dx}[p(x)] q(x) + p(x) \dfrac{d}{dx}[q(x)]$

4. $\dfrac{d}{dx}\left[\dfrac{p(x)}{q(x)}\right] = \dfrac{\dfrac{d}{dx}[p(x)]q(x) - p(x) \dfrac{d}{dx}[q(x)]}{(q(x))^2}$

Here Are Some of the Essential Properties of Derivatives:

(i) $\dfrac{d}{dx}(x^n) = nx^{n-1}$

(ii) $\dfrac{d}{dx}(\sin x) = \cos x$

(iii) $\dfrac{d}{dx}(\cos x) = -\sin x$

(iv) $\dfrac{d}{dx}(\tan x) = \sec^2 x$

(v) $\dfrac{d}{dx}(\cot x) = -\text{cosec}^2\,x$

(iv) $\dfrac{d}{dx}(\sec x) = \sec x \tan x$

(v) $\dfrac{d}{dx}(\text{cosec }x) = -\text{cosec }x \cot x$

(vii) $\dfrac{d}{dx}(a^x) = a^x log_ea$

(ix) $\dfrac{d}{dx}(e^x) = e^x$

(x) $\dfrac{d}{dx}(log_ex) = \dfrac{1}{x}$

A Few Standard Derivatives

Solved Examples On How To Solve Limits

You will come across the following types of limits examples along with step-by-step solutions in the limits question bank chapter provided by Vedantu.

Example: Identify the limit of the following expression?

$\lim_{x \to 5} \dfrac{x^2 - 5}{x^2 + x - 30}$

Solution:

The limit provided is the ratio of two polynomials, $x = 5$. This certainly makes both the numerator as well as the denominator equivalent to zero (0). We are required to factor both the numerator as well as denominator as shown below.

$\lim_{x \to 5} \dfrac{(x - 5)(x + 5)}{(x - 5)(x + 6)}$

Simplify the expression to get:-

$\lim_{x \to 5} \dfrac{x + 5}{x + 6} = \dfrac{10}{11}$

Introduction to Limits by Factoring

Now, this particular method is quite an interesting way of solving limits. In these types of limits, if you try to substitute, you will obtain an indetermination. For example:

$\lim_{x \to 1} x^2{\dfrac{x^2 - 1}{x - 1}}$

Note that if you simply substitute x by 1 in the algebraic equation, you will have 0/0. So, what do you think can be done? We can take the help of our algebraic skills for the purpose of simplifying the expression. In the example quoted previously, we can factor the numerator:

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

You are going to find these types of numerical problems on limits easily whenever you notice a quotient of two polynomials. You could try your hands on this method given that there is an indetermination.