Limits and Derivatives Class 11

Limits and derivatives class 11 covers topics such as intuitive ideas of derivatives, limits, limits of trigonometry functions and derivatives. Limits and derivatives have scope in not only Maths but also they are highly used in Physics to derive some particular derivations. We will discuss here limits and derivatives explained in class 11 syllabus with properties and formulas.

Limits and derivatives class 11 covers the base for class 12th limits and derivatives topics. If you have understood the concepts of limits and derivatives in class 11 then you can easily get through the same topic and related problems in your next level of class.

In limits and derivatives class 11 topic, we will cover introductions of limits and derivatives, formulas of limits and derivatives, properties of limits and derivatives and also some example questions for both the topics. Let us learn and understand here all these topics.

Limits and Derivatives Class 11 Definitions

Definition of Limits:

A limit of a function f(x) is defined as a value, where the function reaches as the limit reaches some value. Limits are used to define integration, integral calculus and continuity of the function.

If f(y) is a function, then the limit of the function can be represented as;

\(\lim_{y\rightarrow c}\)

This is the general expression of limit, where c is any constant value. But there are some important properties of limits which we will discuss here.

Properties of limits of a given function:

A. Let p and q be any two functions and a be any constant value in such a way that, there exists, \(\lim_{ x \rightarrow a} p(x)\) and \(\lim_{ x \rightarrow a} q(x)\). Now, check below the properties of limits.

Limits and Derivatives class 11

B. For any positive integers m,

\(\lim_{x\rightarrow a}\frac{x^m – a^m}{x-a}\) = nam-1

C.Limits of trigonometric functions;

If p and q are real-valued function with the same domain, such that, p(x) ≤ q(x) for all the values of x. For a value b, if both \(\lim_{ x \rightarrow a} p(x)\) and \(\lim_{ x \rightarrow a} q(x)\) exists then,

\(\lim_{x\rightarrow b}p(x)\leq \lim_{x\rightarrow b}q(x)\)

Example:Let f(x) = x2 – 4. Compute \(\lim_{ x \rightarrow 2} f(x)\).

Solution: \(\lim_{ x \rightarrow 2} f(x)\) = \(\lim_{ x \rightarrow 2} x^2 – 4\)

= 22 – 4 = 4 – 4 = 0

Definition of Derivatives:

A derivative is defined as the rate of change of a function or quantity with respect to others. The formula for derivative can be represented in the form of;

\(\lim_{ a \rightarrow 0} \frac{f(x+a)- f(x)}{a}\)

The derivative of a function f(x) is denoted as f’(x). Now, let us see the properties of derivatives.

Properties of derivatives for given functions:

Let for functions p(x) and q(x), the properties of derivatives are;

Limits and Derivatives Class 11-1

Example: Calculate \(\frac{\mathrm{d} (x^4+1)}{\mathrm{d} x}\)

Solution: We know,

\(\frac{\mathrm{d} (x^n)}{\mathrm{d} x}\) = n xn-1 and derivative of a constant value is 0.

Therefore,

\(\frac{\mathrm{d} (x^4+1)}{\mathrm{d} x}\) = 4x3

General Derivative Formulas

\(\frac{\mathrm{d} (x)}{\mathrm{d} x}\)

= 1

\(\frac{\mathrm{d} (ax)}{\mathrm{d} x}\)

= a

\(\frac{\mathrm{d} (x^n)}{\mathrm{d} x}\)

= n xn-1

\(\frac{\mathrm{d} (cos x)}{\mathrm{d} x}\)

= – Sin x

\(\frac{\mathrm{d} (sin x)}{\mathrm{d} x}\)

= Cos x

\(\frac{\mathrm{d} (tan x)}{\mathrm{d} x}\)

= Sec2 x

\(\frac{\mathrm{d} (co tx)}{\mathrm{d} x}\)

= – Cosec2 x

\(\frac{\mathrm{d} (sec x)}{\mathrm{d} x}\)

= Sec x. tan x

\(\frac{\mathrm{d} (cosec x)}{\mathrm{d} x}\)

= – Cosec x . cot x

\(\frac{\mathrm{d} (ln x)}{\mathrm{d} x}\)

= \(\frac{1}{x}\)

\(\frac{\mathrm{d} (e^{x})}{\mathrm{d} x}\)

= ex

\(\frac{\mathrm{d} (a^x)}{\mathrm{d} x}\)

= ax(ln a)

\(\frac{\mathrm{d} (sin^{-1} x)}{\mathrm{d} x}\)

= \(\frac{1}{\sqrt{1-x^2}}\)

\(\frac{\mathrm{d} (tan^{-1} x)}{\mathrm{d} x}\)

= \(\frac{1}{1+x^2}\)

\(\frac{\mathrm{d} (sec^{-1} x)}{\mathrm{d} x}\)

= \(\frac{1}{|x|\sqrt{x^2-1}}\)

Limits and Derivatives class 11 Important Questions and Solutions

1. Question: Evaluate \(\lim_{x\rightarrow 2}[\frac{x^2-4}{x-2}]\).

Solution: \(\lim_{x\rightarrow 2}[\frac{x^2-4}{x-2}]\) = \(\lim_{x\rightarrow 2}\frac{(x+2)(x-2)}{x-2}\)

Cancel the term x-2 from numerator and denominator. Now we get,

\(\lim_{x→ 2} x+2 = 2+2 = 4\)

2. Question: Solve \(\lim_{x\rightarrow 0}\frac{sin 2x}{x}\)

Solution: Given,\(\lim_{x\rightarrow 0}\frac{sin 2x}{x}\)

We can write it as;

\(\lim_{x\rightarrow 0}\frac{sin 2x}{2x}\) × 2

Since, \(\lim_{x\rightarrow 0}\frac{sin x}{x}\) = 1

Therefore, \(\lim_{x\rightarrow 0}\frac{sin 2x}{2x}\) × 2 = 1 × 2 = 2

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limn 20x=1 cos 2n(x10) is equal to

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