Standard Limits - Formula and Solved Examples

Standard limits formulas will help students to do a quick revision before the exam. A limit is a value that a function approaches as the input approaches some value. In this article, we will find the standard limits formulas and some solved examples. The limit of a function is usually denoted by

\begin{array}{l} lim_{x \to a} f (x) = L \end{array}

.

This is read as the limit of f(x) as x tends to c equals L. x → c is read as x tends to c.

Limits of Trigonometry Functions

\begin{array}{l} 1. lim_{x \to 0} \sin x = 0 \end{array}

\begin{array}{l} 2. lim_{x \to 0} \cos x = 1 \end{array}

\begin{array}{l} 3. lim_{x \to 0} \frac{\tan x}{x} = 1 \end{array}

\begin{array}{l} 4. lim_{x \to 0} \frac{\sin x}{x} = 1 \end{array}

\begin{array}{l} 5. lim_{x \to 0} \frac{\sin^{- 1} x}{x} = 1 \end{array}

\begin{array}{l} 6. lim_{x \to 0} \frac{\tan^{- 1} x}{x} = 1 \end{array}

Limits of form 1^∞

\begin{array}{l} 1. lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e \end{array}

\begin{array}{l} 2. lim_{x \to \infty} (1 + \frac{1}{x})^{x} = e \end{array}

\begin{array}{l} 3. lim_{x \to \infty} (1 + \frac{a}{x})^{x} = e^{a} \end{array}

Limits of Log and Exponential Functions

\begin{array}{l} 1. lim_{x \to 0} e^{x} = 1 \end{array}

\begin{array}{l} 2. lim_{x \to 0} \frac{e^{x} - 1}{x} = 1 \end{array}

\begin{array}{l} 3. lim_{x \to 0} \frac{a^{x} - 1}{x} = l o g_{e} a \end{array}

\begin{array}{l} 4. lim_{x \to 0} \frac{\log (1 + x)}{x} = 1 \end{array}

\begin{array}{l} 5. lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1} \end{array}

L’Hospital’s Rule

\begin{array}{l} If lim_{x \to a} \frac{f (x)}{g (x)} is of the form \frac{0}{0}, then lim_{x \to a} \frac{f (x)}{g (x)} = \frac{f^{'} (x)}{g^{'} (x)} . \end{array}

Also, Read:

Limits, Continuity and Differentiability – Evaluations and Examples

Limits Solved Examples

Solved Examples

Example 1: Let [x] be the greatest integer less than or equal to x. Then, at which of the following point(s) the function f(x) = x cos (π(x + [x])) is discontinuous?

(a) x = 2

(b) x = 0

(c) x = 1

(d) x = -1

Solution:

Given f(x) = x cos (π(x + [x]))

At x = 2

lim_{x→ 2-}x cos (π(x + [x])) = 2 cos (π+2π)

= 2 cos 3π

= -2

lim_{x→ 2+}x cos (π(x + [x])) = 2 cos (2π+2π)

= 2 cos 4π

= 2

LHL ≠ RHL

So, f(x) is discontinuous at x = 2.

At x = 0

lim_{x→ 0-}x cos (π(x + [x])) = 0 cos (-π+0)

= 0

lim_{x→ 0+}x cos (π(x + [x])) = 0

LHL = RHL

So, f(x) is continuous at x = 0.

At x = 1

lim_{x→ 1-}x cos (π(x + [x])) = cos (π)

= -1

lim_{x→ 1+}x cos (π(x + [x])) = cos 2π

= 1

LHL ≠ RHL

So, f(x) is discontinuous at x = 1.

At x = -1

lim_{x→ -1-}x cos (π(x + [x])) = -cos (-3π)

= 1

lim_{x→ -1+}x cos (π(x + [x])) = -cos 2π

= -1

LHL ≠ RHL

So, f(x) is discontinuous at x = 1.

The function is discontinuous at x = 2, 1, -1

Hence, options a, c and d are correct.

Example 2: If lim_{x→ 0} ((a-n)nx – tan x) sin nx/x² = 0, where n is non zero real number, then a is equal to

(a) 0

(b) (n+1)/n

(c) n

(d) n + 1/n

Solution:

lim_{x→ 0} ((a-n)nx – tan x) sin nx/x² = 0

=> lim_{x→ 0}(n sin nx/nx)[ (a-n)n – tan x/x] = 0

=> n. [(a-n)n-1] = 0

=> a = n + 1/n

Hence, option d is the answer.

Example 3: Evaluate lim_{x→ 0} (tan x – sin x)/sin³x.

(a) 1

(b) 1/2

(c) 0

(d) 2

Solution:

lim_{x→ 0} (tan x – sin x)/sin³x = lim_{x→ 0} sin x(1/cos x – 1)/sin³x

= lim_{x→ 0}(1-cos x)/cos x sin²x

= lim_{x→ 0}(1-cos x)/cos x (1- cos²x)

= lim_{x→ 0}2 sin²(x/2)/cos x .4 sin²x/2 cos²x/2

= lim_{x→ 0}2/4 cos x cos²x/2

= 1/2

Hence, option b is the answer.

Example 4: lim_x→0|x|/x is equal to

(a) 1

(b) -1

(c) 0

(d) does not exist

Solution:

RHL = lim_x→0+|x|/x = x/x = 1

LHL = lim_x→0-|x|/x = -x/x = -1

So, the limit does not exist.

Hence, option d is the answer.

Example 5: lim_{x→ 3}(x²-9)/(x-3) is

(a) 3

(b) -3

(c) 9

(d) 6

Solution:

lim_{x→ 3}(x²-9)/(x-3) = lim_{x→ 3}(x-3)(x+3)/(x-3)

= lim_{x→ 3}(x+3)

= 3+3

= 6

Hence, option d is the answer.

Related Videos

Limits, Continuity and Differentiability – JEE Questions

76,462

Limits, Continuity and Differentiability – JEE Main Questions

2,608

Frequently Asked Questions

Q1

Give the L’Hospital’s rule of limits.

If lim_x→af(x)/g(x) is of the form 0/0, then we find the derivative of the numerator and the derivative of the denominator and find the limits.

lim_x→af(x)/g(x) = lim_x→af’(x)/g’(x).

Q2

What is the value of lim_x→0e^x?

The value of lim_x→0e^x = 1.

Q3

What is the value of lim_x→0 cos x?

The value of lim_x→0 cos x = 1.

Comments

Leave a Comment Cancel reply