Algebra of Limits

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 limxpA(x)\lim_{x \rightarrow p} A(x) and limxpB(x)\lim_{x \rightarrow p} B(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 limxpA(x)\lim_{x \rightarrow p} A(x) and limxpB(x)\lim_{x \rightarrow p} B(x) exists, then

limxpA(x)±limxpB(x)=limxp(A±B)(x)\lim_{x \rightarrow p} A(x) \pm \lim_{x \rightarrow p} B(x)= \lim_{x \rightarrow p} (A \pm B)(x)

Important Rules:

Let A(x) and B(x) are function of x such that limxpA(x)=l\lim_{x \rightarrow p} A(x)=l and limxpB(x)=m\lim_{x \rightarrow p} B(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.

limxp(A±B)(x)=limxpA(x)±limxpB(x)=l±m\lim_{x \rightarrow p}(A \pm B)(x)= \lim_{x \rightarrow p} A(x) \pm \lim_{x \rightarrow p} B(x) = l \pm m

(ii) The limits of the product of any two given function is same as the product of the limits of that given functions.

limxp(A×B)(x)=limxpA(x)×limxpB(x)=l×m\lim_{x \rightarrow p}(A \times B)(x)= \lim_{x \rightarrow p} A(x) \times \lim_{x \rightarrow p} B(x) = l \times 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.

Limit Rules

(iv) The limit of a constant multiple of the function f(x) is equal to the c times the limit of that function.

limxpkA(x)=klimxpA(x)\lim_{x \rightarrow p}k A(x)= k \lim_{x \rightarrow p}A(x); where k is a constant.

(v) limxpA(x)=limxpA(x)=l\lim_{x \rightarrow p}| A(x)|=|\lim_{x \rightarrow p} A(x)|=|l|

(vi) limxpA(x)B(x)=limxpA(x)limxpB(x)=lm\lim_{x \rightarrow p} A(x)^{B(x)}= \lim_{x \rightarrow p} A(x)^{\lim_{x \rightarrow p} B(x)} = l^m

(vii) limxpAoB(x)=A(limxp\lim_{x \rightarrow p} AoB(x) = A(\lim_{x \rightarrow p} B(x)) = A(m), only if A(x) is continuous at B(x) = m

(viii) limxp=or\lim_{x \rightarrow p}= \infty or – \infty then limxp1A(x)=0\lim_{x \rightarrow p} \frac{1}{A(x)}= 0

Important Points to Remember:

(i) limxp\lim_{x \rightarrow p}(A.B)(x) exists then we have below cases:

(a) both limxp\lim_{x \rightarrow p}A(x) and limxp\lim_{x \rightarrow p}B(x) exists

(b) limxp\lim_{x \rightarrow p}A(x) exists, but limxp\lim_{x \rightarrow p}B(x) does not exist.

Example: A(x) = x and B(x) = 1/x then limxp\lim_{x \rightarrow p}A(x) = 0(exists) and limxp\lim_{x \rightarrow p}B(x) does not exist, but limxp\lim_{x \rightarrow p}(A.B)(x) = 1(exists)

(c) both limxp\lim_{x \rightarrow p}A(x) and limxp\lim_{x \rightarrow p}B(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

limxp\lim_{x \rightarrow p}A(x) does not exist and limxp\lim_{x \rightarrow p}B(x) does not exist, but limxp\lim_{x \rightarrow p}(A.B)(x) = 2(exists).

(ii) If limxp\lim_{x \rightarrow p}[A(x)+B(x)] exists then we have below cases:

(a) If limxp\lim_{x \rightarrow p}A(x) exists, then limxp\lim_{x \rightarrow p}B(x) must exist

Proof: This is true as B = [A + B] – A.

So by the limit theorem, limxp\lim_{x \rightarrow p}B(x) = limxp\lim_{x \rightarrow p}[A(x) + B(x)] – limxp\lim_{x \rightarrow p}B(x) which exists.

(b) Both limxp\lim_{x \rightarrow p}A(x) and limxp\lim_{x \rightarrow p}B(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 limxp\lim_{x \rightarrow p}A(x) and limxp\lim_{x \rightarrow p}B(x) does not exist but limxp\lim_{x \rightarrow p}[A(x) + B(x)] exists.

Methods for Evaluating the Limits

(1) Direct substitution method:

(a). limxp\lim_{x \rightarrow p}f(x)

(b). limxp\lim_{x \rightarrow p} (A(x))/(B(x))

If f(a) and (A(a))/(B(a))both exists and B(a)≠0 then limxp\lim_{x \rightarrow p}f(x) = f(a) and limxp\lim_{x \rightarrow 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:

Some standards limits are:

  • limxaxnanxa=n(a)n1\lim_{x \to a} \frac{x^{n}-a^{n}}{x-a}= n(a)^{n-1}
  • limx0sinx=0\lim_{x \to 0} sin x = 0
  • limx0cosx=1\lim_{x \to 0} cos x = 1
  • limx01cosxx=0\lim_{x \to 0} \frac{1-cos x}{x} = 0
  • limx0sin1xx=1\lim_{x \to 0} \frac{sin^{-1}x}{x} = 1
  • limx0tan1xx=1\lim_{x \to 0} \frac{tan^{-1} x}{x} = 1
  • limx0sinxx=1\lim_{x \to 0} \frac{sin x}{x} = 1
  • limx0tanxx=1\lim_{x \to 0} \frac{tan x}{x} = 1
  • limx0ex=1\lim_{x \to 0} e^{x}=1
  • limx0ex1x=1\lim_{x \to 0} \frac{e^{x}-1}{x}=1
  • limx(1+1x)x=e\lim_{x \to \infty } \left ( 1+\frac{1}{x} \right )^{x}=e
  • limx(1+ax)x=ea\lim_{x \to \infty } \left ( 1+\frac{a}{x} \right )^{x}=e^{a}
  • limx0(1+x)1x=e\lim_{x \to 0} (1+x)^{\frac{1}{x}}=e
  • limx0ax1x=logea\lim_{x \to 0} \frac{a^{x}-1}{x}=log_{e}a
  • limx0log(1+x)x=1\lim_{x \to 0} \frac{log(1+x)}{x}=1

Also Read

Limits at Infinity and Infinite limits

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+\lim_{x \rightarrow +\infty} 1/x = 0 and limx\lim_{x \rightarrow \infty} 1/x2 = 0

So we have: limx+\lim_{x \rightarrow +\infty}f(x) = limy0\lim_{y \rightarrow 0}f(1/y).

Example: limx5x2+17x212x3\lim_{x \rightarrow \infty} \frac{\sqrt{5x^2+1}-\sqrt{7x^2-1}}{2x – 3}

Algebra of Vectors

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

limxa\lim_{x \rightarrow a}f(x) = limxa\lim_{x \rightarrow a}h(x) = L ⇒ limxa\lim_{x \rightarrow a}g(x) = L.

Example: Compute limxx5sinx5x7\lim_{x \rightarrow \infty}\frac{x – 5 sin x}{-5x – 7}


We know that range of sin x is [-1, 1], so min(sin x) = -1 and max(sin x) = 1

⇒ -1 ≤ sin x ≤ 1

⇒ -5 ≤ 5sin x ≤ 5

⇒ 5≥-5.sin x≥-5

⇒ x + 5≥x-5.sin x≥x – 5

x5sinx5x7x5sinx5x7x55x7\frac{x – 5 sin x}{-5x – 7} \leq \frac{x – 5 sin x}{-5x – 7} \leq \frac{x – 5}{-5x – 7} limx(x+5)5x7\lim_{x \rightarrow \infty}\frac{(x + 5)}{-5x – 7}

= -1/5

So we can say that limxx5sinx5x7\lim_{x \rightarrow \infty}\frac{x – 5 sin x}{-5x – 7} is -1/5.

Practice Problems:

  • Compute limx0sin(2+x)sin(2x)x\lim_{x \to 0} \frac{sin(2+x) – sin (2-x)}{x}
  • Prove that limx1(6x+1)=7byusingtheprecisedefinitionofalimit.\lim_{x \to 1} (6x+1) = 7 by using the precise definition of a limit.
  • Find limx0sin(a+x)sin(ax)x\lim_{x\rightarrow0}\frac{sin(a+x)-sin(a-x)}{x}