Limits and Derivatives

Limits

Limits: You and your friends decide to meet at some place outside. Is it necessary that all your friends are living in the same place and walk on the same road to reach that place? No, not always. All friends come from different parts of the city or country to meet at that one single place.

It looks like a convergence of different elements to a single point. Mathematically, it is like a convergence of a function to a particular value. It is an example of limits. Limits show how some functions are bounded. The function tends to some value when its limit approaches some value.

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Introduction to Limits

Suppose we have a function f(x). The value, a function attains, as the variable x approaches a particular value say a, i.e., x → a is called its limit. Here, ‘a’ is some pre-assigned value. It is denoted as

limx→af(x) = l

  • The expected value of the function shown by the points to the left of a point ‘a’ is the left-hand limit of the function at that point. It is denoted as limx→a−  f(x).
  • The points to the right of a point ‘a’ which shows the value of the function is the right-hand limit of the function at that point. It is denoted as limx→af(x).

Limits of functions at a point are the common and coincidence value of the left and right-hand limits.

Limits

The value of a limit of a function f(x) at a point a i.e., f(a) may vary from the value of f(x) at ‘a’.

Properties of Limits

  • limx→a c = c, where c is a constant quantity.
  • The value of limx→a x = a
  • Value of limx→a bx + c = ba + c
  • limx→a xn = an, if n is a positive integer.
  • Value of limx→0+ 1/xr = +∞.
  • limx→0−  1/xr = −∞, if r is odd, and
  • limx→0−  1/xr = +∞, if r is even.

Algebra of Limits

Let p and q be two functions such that their limits limx→a p(x) and limx→a q(x) exist.

  • Limit of the sum of two functions is the sum of the limits of the functions.

limx→a [p(x) + q(x)] = limx→a p(x) + limx→a q(x).

  • Limit of the difference of two functions is the difference of the limits of the functions.

limx→a [p(x) − q(x)] = limx→a p(x) − limx→a q(x).

  • Limit of product of two functions is the product of the limits of the functions.

limx→a [p(x) × q(x)] = [limx→a p(x)] × [limx→a q(x)].

  • Limit of quotient of two functions is the quotient of the limits of the functions.

limx→a [p(x) ÷ q(x)] = [limx→a p(x)] ÷ [limx→a q(x)].

  • Limit of product of a function p(x) with a constant, q(x) = α is α times the limit of p(x).

limx→a [α.p(x))] = α. limx→a p(x).

Limit of Polynomial Function

Consider a polynomial function, f(x) = a0 + a1x + a2x2 + … + anxn. Here, a0, a1, … , an are all constants. At any point x = a, the limit of this polynomial function is

limx→a f(x) = limx→a [a0 +a1x + a2x2 + … + anxn]

= limx→a a0 +a1 limx→a x + a2 limx→a x2 + … + an limx→a xn

or, limx→a = a0 +a1a + a2a2 + … + anan = f(a).

Limit of Rational Function

The limit of any rational function of the type p(x) / q(x), where q(x) ≠ 0 and p(x) and q(x) are polynomial functions, is

limx→a [p(x) / q(x)] = [limx→a p(x)] / [limx→a q(x)] = p(a)/q(a).

The very first step to find the limit of a rational function is to check if it is reduced to the form 0/0 at some point. If it is so, then we need to do some adjustment so that one can calculate the value of the limit. This can be done by

  • Canceling the factor which causes the limit to be of the form 0/0.

Assume a function, f(x) = (x2 + 4x + 4) / (x2 − 4). Taking limit over it for x = −2, the function is of the form 0/0,

limx→−2 f(x) = limx→−2 [(x2 + 4x + 4) / (x2 – 4)]

= limx→−2  [(x +2)2 / (x − 2)(x + 2)] = limx→−2  (x + 2) / (x – 2) = 0/−4 (≠ 0/0) = 0.

  •  Applying the L – Hospital’s Rule.

Differentiating both the numerator and the denominator of the rational function until the value of limit is not of the form 0/0. Assume a function, f(x) = sin x/x. Taking limit over it for x = 0, the function is of the form 0/0.

Taking the differentiation of both sin x and x with respect to x in the limit, limx→0  sin x/x  reduces to limx→0 cos x / 1 = 1. (cos 0 = 1)

Solved Examples for You

Question 1:  Find the limit of limx→2 [x3 + 2x2 + 4x – 2].

Solution:
limx→2 [x3 + 2x2 + 4x – 2] = limx→2 x3 + 2 limx→2 x2 + 4 limx→2 x – limx→2  2
= 23 + 2×22 + 4×2 − 2 = 22.

Question 2:  Find the limit of limx→0 (sin 4x/tan x).

Solution:
limx→0 (sin 4x / tan x) = limx→0 [(sin 4x / 4x) × (4x / tan x)]  (multiplying and dividing by 4x)
or, limx→0 (sin 4x / tan x) = limx→0 (sin 4x / 4x) × 4 limx→0 (x / tan x )
= 1 × 4 = 4 (by using L-Hospital’s Rule).

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