# 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 $\lim_{x \rightarrow p} A(x)$ and $\lim_{x \rightarrow p} B(x)$ 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 $\lim_{x \rightarrow p} A(x)$ and $\lim_{x \rightarrow p} B(x)$ exists, then

$\lim_{x \rightarrow p} A(x) \pm \lim_{x \rightarrow p} B(x)= \lim_{x \rightarrow p} (A \pm B)(x)$

Important Rules:

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

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.

$\lim_{x \rightarrow p}(A \pm B)(x)= \lim_{x \rightarrow p} A(x) \pm \lim_{x \rightarrow p} B(x) = l \pm m$

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

$\lim_{x \rightarrow p}(A \times B)(x)= \lim_{x \rightarrow p} A(x) \times \lim_{x \rightarrow p} B(x) = l \times m$

(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.

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

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

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

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

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

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

Important Points to Remember:

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

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

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

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

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

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

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

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

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

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

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

## Methods for Evaluating the Limits

(1) Direct substitution method:

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

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

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

• $\lim_{x \to a} \frac{x^{n}-a^{n}}{x-a}= n(a)^{n-1}$
• $\lim_{x \to 0} sin x = 0$
• $\lim_{x \to 0} cos x = 1$
• $\lim_{x \to 0} \frac{1-cos x}{x} = 0$
• $\lim_{x \to 0} \frac{sin^{-1}x}{x} = 1$
• $\lim_{x \to 0} \frac{tan^{-1} x}{x} = 1$
• $\lim_{x \to 0} \frac{sin x}{x} = 1$
• $\lim_{x \to 0} \frac{tan x}{x} = 1$
• $\lim_{x \to 0} e^{x}=1$
• $\lim_{x \to 0} \frac{e^{x}-1}{x}=1$
• $\lim_{x \to \infty } \left ( 1+\frac{1}{x} \right )^{x}=e$
• $\lim_{x \to \infty } \left ( 1+\frac{a}{x} \right )^{x}=e^{a}$
• $\lim_{x \to 0} (1+x)^{\frac{1}{x}}=e$
• $\lim_{x \to 0} \frac{a^{x}-1}{x}=log_{e}a$
• $\lim_{x \to 0} \frac{log(1+x)}{x}=1$

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

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

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

## 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

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

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

Solution:

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

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

= -1/5

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

### Practice Problems:

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