Standard Limits

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

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

Limits of form 1^∞

Limits of Log and Exponential Functions

L’Hospital’s Rule

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.

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