The solution of equations and sets of equations is an essential and historically important part of what we call Algebra. Algebra is commonly used in formulas when we can change one of the numbers or at least one of the numbers is unknown. In this article, you will learn the algebra of limits.
Algebra methods are used to evaluate the limits. Some of the important methods are factorization method, evaluation using standard limits, direct substitution method, rationalization and evaluation of limits at infinity.
Let A and B be two functions such that their limits limx→pA(x) and limx→pB(x) exists, then the limit of the sum of two functions is the sum of the limits of the functions.
Limit of Sum of Two Functions
Let A and B be two functions such that their limits limx→pA(x) and limx→pB(x) exists, then
limx→pA(x)±limx→pB(x)=limx→p(A±B)(x)
Important Rules:
Let A(x) and B(x) are function of x such that limx→pA(x)=l and limx→pB(x)=m
If l and m exist, then
(i) The sum/subtraction of the limits of two given functions if equal to the limit of the sum/subtraction of two functions.
limx→p(A±B)(x)=limx→pA(x)±limx→pB(x)=l±m
(ii) The limits of the product of any two given function is same as the product of the limits of that given functions.
limx→p(A×B)(x)=limx→pA(x)×limx→pB(x)=l×m
(iii) If we have the quotient of two functions then the limit that term is same as the quotient of their limits included the condition that the limit of the denominator is not equal to zero.
(iv) The limit of a constant multiple of the function f(x) is equal to the c times the limit of that function.
limx→pkA(x)=klimx→pA(x); where k is a constant.
(v) limx→p∣A(x)∣=∣limx→pA(x)∣=∣l∣
(vi) limx→pA(x)B(x)=limx→pA(x)limx→pB(x)=lm
(vii) limx→pAoB(x)=A(limx→p B(x)) = A(m), only if A(x) is continuous at B(x) = m
(viii) limx→p=∞or–∞ then limx→pA(x)1=0
Important Points to Remember:
(i) limx→p(A.B)(x)
⇒ both limx→pA(x) and limx→pB(x) exists
(ii) limx→p(A.B)(x)
⇒ limx→pA(x) exists, but limx→pB(x) does not exist.
Example: A(x) = x and B(x) = 1/x then limx→pA(x) = 0(exists) and limx→pB(x) does not exist, but limx→p(A.B)(x) = 1(exists)
(iii) limx→p(A.B)(x)
⇒ both limx→pA(x) and limx→pB(x) does not exist
Example: A(x) = 1 ∀ x ≥ 0 and 2 ∀ x < 0 and B(x) = 2 ∀ x ≥ 0 and 1 ∀ x < 0 then
limx→pA(x) does not exist and limx→pB(x) does not exist, but limx→p(A.B)(x) = 2(exists).
(iv) If limx→p[A(x)+B(x)] exists then we have three cases:
(a) If limx→pA(x) exists, then limx→pB(x) must exist
Proof: This is true as B = [A + B] – A.
So by the limit theorem, limx→pB(x) = limx→p[A(x) + B(x)] – limx→pB(x) which exists.
(b) Both limx→pA(x) and limx→pB(x) does not exist
Example: A(x) = [x], where [.] represents greatest integer function.
B(x) = {x}, where {.} represents fraction part function.
Here both the limits limx→pA(x) and limx→pB(x) does not exist but limx→p[A(x) + B(x)] exists.
Methods for Evaluating the Limits
(1) Direct substitution method:
(a). limx→pf(x)
(b). limx→p (A(x))/(B(x))
If f(a) and (A(a))/(B(a))both exists and B(a)≠0 then limx→pf(x) = f(a) and limx→p (A(x))/(B(x))=(A(a))/(B(a)).
(2) Factorisation and rationalisation: Do the factorisation and rationalisation whenever needed.
(3) Evaluation of limits using standard substitution:
The Limit of a function shows the behavior of the function at a particular point. The limit is the output value of the function for which the input value approaches closer and closer to a particular point. Lets see with the help of an example how to compute infinity and infinite limits.
We know that limx→+∞ 1/x = 0 and limx→∞ 1/x^2 = 0
So we have: limx→+∞f(x) = limy→0f(1/y).
Example: limx→∞2x–35x2+1−7x2−1
Sandwich Theorem
Sometimes we use the squeeze principle on limits where usual algebraic methods are not effective. In this method, adjust the given problem in between two other simpler functions whose limit can easily be computed and equal. We need knowledge of algebra and inequalities.
If f, g and h are three functions such that f(x) < g(x) < h(x) for all x in some interval containing the point x = a, and if
limx→af(x) = limx→ah(x) = L ⇒ limx→ag(x) = L.
Example: Compute limx→∞−5x–7x–5sinx
Solution:
We know that range of sin x is [-1, 1], so min(sin x) = -1 and max(sin x) = 1