What is Evaluating?
Basically, evaluating is the general term for solving systematic determination by using a set of standards set for that system.
In mathematics, evaluation means to E-Value-Ate. It means to solve equations, functions, or expressions to get a numerical expression of it.
Evaluating Limits:
Why are you worried about how to solve limits?
Limits can be solved into numerical values by using many techniques or sets of standards. These techniques include simple substitutions, algebraic substitution, conjugation, or recognizing a pattern etc.
In fact, there are many evaluating patterns to get an accurate answer . Let’s see some of the important methods of solving limits here:
Simple substitution:
The first approach in the concept of evaluating limits is to simply substitute the value of limit in a function. Let’s see how to solve limits by using this method:
Evaluate limit:
$$\lim\limits_{x \to 3}{5x-2}$$
Solution :
In a simple substitution method of solving limits, we have only replaced x by its limiting value.
Here,
Replace x by 3,
$$\lim\limits_{x \to 3}{5x-2} = {5(3)-2}=0$$
Hence,
$$\lim\limits_{x \to 3}{5x-2}={13}$$
Now you can see how to solve limits by using simple method of evaluating limits. It’s quite an easy to replace the value of the function of limit
Let’s try another example
$$\lim\limits_{x \to 3} \frac{x^2-9}{x-3}$$
Again, simply replace x by 3
$$\lim\limits_{x \to 3} \frac{x^2-9}{x-3} = \frac{(3)^2-9}{3-3} = \frac{0}{0}$$
Now, 0/0 not looks like a great practice. We have to try some other evaluating technique here.
Algebraic substitution of Evaluating Limits:
In algebraic substitution, we have to first factorize or substitute the algebraic equation to its simplest form. Let’s try
We know $${x^2-y^2}={(x+y)(x-y)}$$
In our given function,
$$x^2-9=(x+3)(x-3)$$
Now, by replacing we will get
$$\lim\limits_{x \to 3} \frac{x^2-9}{x-3} = \frac{(x+3)(x-3)}{(x-3)} = {x+3} $$ $$\lim\limits_{x \to 3} \frac{x^2-9}{x-3} = {x+3} $$
The function is in the simplest form. Now use the simplest substitution method here
$$\lim\limits_{x \to 3} \frac{x^2-9}{x-3} = {3+3} = {6} $$
Now, it’s an appropriate answer.
Recognizing a pattern:
Evaluate the limit of the given sequence:
$$\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},{......}$$
In a given sequence, the fraction gets larger than the previous fraction and we can recognize that numerator is always one less than by its denominator. So, the sequence approaches closer and closer to 1.
In this way limit of the sequence is 1.
Conjugate method of Solving Limits:
In this method, the first question is what is conjugate?
So, the conjugate is formed by simply changing the sign used in a binomial expression to form opposite signed binomial.
E.g
- The conjugate of x+y is x-y.
- The conjugate of 3+x is 3-x.
Or vice versa
Sometimes when simple substitution and algebraic expression method is not enough. We have to use the conjugate method for evaluating limits of 0/0 form.
In a conjugate limits, we have to multiply numerator and denominator by conjugate used in a function.
Let’s understand it by an example
Example:
$$\text{Evaluate:} \;\;\;\;\;\;\;\;\;\;\; \lim\limits_{x \to 0} \frac{\sqrt {x+a} - \sqrt a}{x} $$
solution:
By using simple substitution, we have 00 at x=0.
So, rationalize the numerator by using the conjugate method.
Therefore,
$$\lim\limits_{x \to 0} \frac{\sqrt {x+a} - \sqrt a}{x} = \lim\limits_{x \to 0} \left( \frac{\sqrt {x+a} - \sqrt a}{x} \right) \left( \frac{\sqrt {x+a} + \sqrt a}{\sqrt {x+a} + \sqrt a} \right)$$ $$ \lim\limits_{x \to 0} \frac{\sqrt {x+a} - \sqrt a}{x} = \lim\limits_{x \to 0} \left( \frac {(\sqrt {x+a})^2 - (\sqrt a)^2} {x(\sqrt {x+a} + \sqrt a)} \right) $$ $$ \lim\limits_{x \to 0} \frac{\sqrt {x+a} - \sqrt a}{x} = \lim\limits_{x \to 0} \frac{x+a-a}{x(\sqrt {x+a} + \sqrt a)} $$ $$ \Rightarrow \lim\limits_{x \to 0} \frac{\sqrt {x+a} - \sqrt a}{x} = \lim\limits_{x \to 0} \frac{1}{\sqrt {x+a} + \sqrt a} $$
Now, replace x by 0(simple substitution)
$$ \Rightarrow \lim\limits_{x \to 0} \frac{\sqrt {x+a} - \sqrt a}{x} = \frac{1}{\sqrt {0+a} + \sqrt a} $$ $$ \Rightarrow \lim\limits_{x \to 0} \frac{\sqrt {x+a} - \sqrt a}{x} = \frac{1}{2 \sqrt {a}} $$
In this way, you can resolve undefined conjugate limits by using the conjugate method.
Evaluating Limits at Infinity:
We know it’s not possible to directly answer that how to solve limits at infinity. In this case, we divide both numerator and denominator by the highest power of x which is appeared in the denominator, and then using general theorem of evaluating limits here. The general theorem for solving limits are as follow:
$$ \lim\limits_{x \to \infty} \frac{1}{x} = 0$$
Let’s see this method of evaluating limits at infinity using an example.
$$ \text{Evaluate} \;\;\;\;\;\;\;\;\;\;\; \lim\limits_{x \to \infty} \frac{4x^4-5x^3}{3x^5+2x^2+1} $$
Solution:
$$ \text{Divide numerator and denominator by } x^5, \text{we will get}$$ $$ \lim\limits_{x \to \infty} \frac{4x^4-5x^3}{3x^5+2x^2+1} = \lim\limits_{x \to \infty} \frac{ \frac{4}{x} - \frac{5}{x^2} }{3 + \frac{2}{x^3} + \frac{1}{x^5}} $$
Now apply the general theorem of evaluating limits in it
$$ \lim\limits_{x \to \infty} \frac{4x^4-5x^3}{3x^5+2x^2+1} = \frac{0-0}{3+0+0} = 0 $$
So, the limit at given function is:
$$ \lim\limits_{x \to \infty} \frac{4x^4-5x^3}{3x^5+2x^2+1} = 0 $$
Here, we learned different methods for evaluating limits and also learned how to solve limits approaching infinity. It's quite simple to evaluating limits at infinity