Introduction to Limits
In mathematics, a limit is defined as a value approaching the output of the given input values by a function. In calculus and mathematical analysis, the limit is important and it is used to describe integrals, derivatives, and continuity. It is used in the research process and is often concerned with the function’s behavior. Let assume, you and your friends decide to meet outside somewhere. Do you have to make all your friends live in the same place and walk the same way to this site? No not just now and then. Your all friends come to meet in one place from various parts of the town or world. Assume we have an f(x) function. The value, a function, arrives with the variable x reaching a given value, meaning that the variable an is referred to as its limit. Here, ‘a is an initial value. It’s labeled as
lim x ⇢ a f(x) = 1
The predicted function value indicated by point ‘a’ on the left is the left-hand limit of this function. It is called
lim x ⇢ -a f(x) = 1
The predicted function value indicated by point ‘a’ on the right is the right-hand limit of this function. It is called
lim x ⇢ +a f(x) = 1
Let us consider a real-valued function “f” and the real number “c”, the limit is normally defined as
lim x ⇢ p f(x) = 1
It is read as “the limit of f of x, as x approaches p equals L”. The “lim” shows the limit, and fact that function f(x) approaches the limit L as x approaches p is described by the right arrow.
Properties of Limits
- lim x ⇢ a k = k, where k is a constant quantity
- The value of lim x ⇢ a x = a
- Value of lim x ⇢ a bx + c = ba + c
- lim x ⇢ a xn = an if n is a positive integer.
- Value of lim x ⇢ +0 1/xr = +∞
- lim x ⇢ −0 1/xr = −∞, if r is odd
- lim x ⇢ −0 1/xr = +∞, if r is even
Algebra of Limit
Let m and n be two functions such that their limits
lim x ⇢ a m(x) and lim x ⇢a n(x) exists.
- The limit of the sum of two functions is the sum of the limits of the functions.
lim x ⇢ a [m(x) + n(x)] = lim x ⇢ a m(x) + lim x ⇢ a n(x)
- The limit of the difference of two functions is the difference between the limits of the functions.
lim x ⇢ a [m(x) – n(x)] = lim x ⇢ a m(x) – lim x ⇢ a n(x)
- The limit of product of two functions is the product of the limits of the functions.
lim x ⇢ a [m(x) × n(x)] = lim x ⇢ a m(x) × lim x ⇢ a n(x)
- The limit of quotient of two functions is the quotient of the limits of the functions.
lim x ⇢ a [m(x) ÷ n(x)] = lim x ⇢ a m(x) ÷ lim x ⇢ a n(x)
- The limit of product of a function m(x) with a constant, n(x) = α is α times the limit of m(x).
lim x ⇢ a [α.m(x)] = α.lim x ⇢ a m(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
lim x ⇢ a f(x) = lim x ⇢ a [a0 + a1x + a2x2 + . . . + anxn] = lim x ⇢ a a0 + a1lim x ⇢ a x + a2lim x ⇢ a x2 + . . . + anlim x ⇢ a xn
lim x ⇢ a a0 + a1x + a2x2 + . . . + anxn = f(a)
Limit of Rational Function
The limit of any rational function of the type m(x)/n(x), where n(x) ≠ 0 and m(x) and n(x) are polynomial functions, is:
lim x ⇢ a [m(x)/n(x)] = lim x ⇢ a m(x)/lim x ⇢ a n(x) = m(a)/m(b)
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 adjustments 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,
lim x ⇢ -2 f(x) = lim x ⇢ 0 [(x2 + 4x + 4)]
= lim x ⇢ -2 [( x + 2)2/(x – 2)( x – 2)]
= lim x ⇢ -2 [(x – 2)/(x – 2)]
= 0/-3 ( ≠ 0/0 ) = 0
Applying the L – Hospital’s Rule
Differentiating both the numerator and the denominator of the rational function until the value of the 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, lim x ⇢ 0 sin x/x reduces to lim x ⇢ 0 cos x/1 = 1.(cos 0 = 1)
Question 1: lim x ⇢ 6 x/3
This can be easily done using the substitution method lim x ⇢ 6 x/3 = 6/3 = 2
Question 2: lim x ⇢ 2 x2 – 4/x2 – 2
x2 – 4 can factorized in (x2 – 22) = ( x – 2 )( x – 2 )
= lim x ⇢ 2 x2 – 4/x2 – 2
= lim x ⇢ 2 (x- 2)(x – 2)/x – 2 = lim x ⇢ 2 x + 2/1
Question 3: lim x ⇢ 1/2 2x – 1/4x2 – 1
4x2 – 1 can be factorized in a (2x2) – (12) = (2x + 1) (2x – 1)
= lim x ⇢ 1/2 2x – 1/4x2 – 1
= lim x ⇢ 1/2 2x- 1/(2x – 1) (2x + 1)
= lim x ⇢ 1/ 2 1/(2x + 1)
= 1/2 × (1/2) + 1
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