Algebra of limits Theorem Proof

The algebra of limits theorem is used to calculate the limit of any algebraic expression. It is used widely in mathematics. Limit of any algebraic expression f(x), on a specific value a, is equated to, which is x→a is demonstrated as, 

    Lim x→a f(x) = l

This statement is only reliable when the functions which lie in the left of the expression are the left-hand limit of the whole expression. The left-hand limit is demonstrated as a, x→a−  f(x). The right-hand limit is demonstrated as a  x→a+ f(x). In the Algebra of limits theorem, compute these two limits to get the value of an algebraic expression. 

Algebra of limits 

Algebra of limits is the systematic process to know the value of variables in the algebraic expressions with the help of limits. If the right-hand limit is equal to the left-hand limit, then it is said that the function is defined, and the value on which the function gets defined is the value of the expression. It means the value which we get after computing the limit is the zero of the algebraic expression. If we put that value in the equation, either we get the zero, or the same quantity is written in equals the equation.

Algebra of limits examples

Some Algebra of limits examples;

1. Algebra of limits example of the limit of A Polynomial Function

Solve: x→alim (x3 – x2 + 1) 

Solution:  

Put the value of x = 1 in the given equation. 

x→alim (x3 – x2 + 1) = x→1lim (13 – 12 + 1)

x→alim (x3 – x2 + 1) = 1 – 1 + 1

x→alim (x3 – x2 + 1) = 1

2. Algebra of limits example of a Rational Function By Direct Substitution.

Solve:  x→1lim = (x2 + 1)/ (x + 100) 

Solution:

Put the value of x = 1 in the given equation. 

x→1lim = x2 + 1/ x + 100 =  12 + 1/ 1 + 100

        = 1 + 1 / 1 + 100

       =  2/101

Hence, x→1lim = x2 + 1/ x + 100 = 2/101

Algebra of limits theorem

The Algebra of limits theorem demonstrates that if 

       x→alim f(x) = l

And, x→alim g(x) = m

Then, 

         x→alim f(x).g(x) = l . m

Algebra of limits theorem proof

Let’s understand Algebra of limits theorem proof:

Given, 

         x→alim f(x) = l

 Or,    x→alim g(x) = m

To proof: x→alim f(x).g(x) = l .m

Proof: 

Step 1:  x→alim |f(x).g(x) – lm | = |f(x).g(x) – f(x).m + f(x).m – lm|

Step 2:   x→alim |f(x).g(x) – lm | = |f(x).(g(x) – m) + m (f(x) – l)| 

Step 3: x→alim |f(x).g(x) – lm | ≤ |f(x)||(g(x) – m)|+ |m| |(f(x) – l|          …Equation (1)

Since, x→alim f(x) = l, where l→ 0, then f(x) → 0 in 0 < |x-a|< δ      

For some δ > 0

Therefore, f(x) is bounded in a. 

Hence, there exist k > 0, in a way that, |f(x)| < k,…….. Equation (2)

Whenever, 0 < |x – a|<δ

For some δ2 < 0

Let, E > 0 be given, 

Since, 

      x→alim f(x) = l and, 

       x→alim g(x) = m, 

In a way that, δ2 > 0.

|f(x) – l| = E/ 2(1 + |m|)  …. Equation (3).

(This is for 0 < |x – a| < δ2)  

And, 

|g(x) – m| = E/ 2k    …… Equation (4).

(This is for 0 < |x – a| < §3) 

If we choose § = min{δ1 , δ2 , δ3}, 

Then, 

Algebraic properties of limits 

So, equation (1),equation (2) and equation(3) hold good. 

For 0 < |x – a| < δ

Therefore, from equation (1),

|f(x).g(x) – lm | < k. E/ 2k + |m|. E/ 2(1 + |m|)

(This is for, 0 < |x – a| < §) 

Now , E/2 + E/2= E

Hence, according to the definition, 

   x→alim f(x).g(x) = l.m = x→alimf(x). x→alim g(x)

Algebraic properties of limits

Let’s learn about the algebraic properties of limits. 

• Limit of any constant in algebraic expressions remains constant. 

• limx→a cx + d = ca + d

• lim x→a x = a

• Lim x→a xn = an, in this equation, n is a positive integer 

• Lim x→0+  1/xr= +∞, here the value of r is even. 

• Lim x→0−  1/xr = −∞, here the value of r is odd. 

Important identities 

Some Important identities are:

• limy→a m(y) + limy→a  n(y) = limy→a (m(y)+n(y))

• limy→a  m(y)−limy→a n(y)=limy→a (m(y)−n(y))

• limy→a  m(y).limy→a n(y)=limy→a m(y). n(y)

•  limy→a m(y)/limy→a n(y)=limy→a m(y)/ n(y)  

• Where limy→a n(y)≠0

Methods for evaluating the algebra of limits:

Some Important methods are:

• Factorization

• Direct Substitution

• Evaluation of Algebraic limits at infinity

• Rationalization Method

Conclusion

Algebra of limits theorem is used frequently for computing the values of x in the algebraic expressions by using the method of limits. There are also many different ways to evaluate the value of algebraic expression by limit, but this method is used extensively. The Algebra of limits theorem method is used after the method in which the left-hand and right-hand limits are found. Along with this, the algebraic expressions properties of limits helps in finding their value instantly. The identities of algebraic expressions also help in computing the large and difficult questions of algebraic expressions and algebra of limits examples.