In mathematics, limits is one the major concepts of calculus and can be applied to different types of functions. Application of limits to the given functions results in another function and sometimes produces the result as 0. In this article, you will learn how to apply limits for polynomials and rational functions along with solved examples.
Limits of Polynomials
A function f is said to be a polynomial function of degree n,
\(\begin{array}{l}f(x)=a_{0}+a_{1}x+a_{2}x^{2}+….+a_{n}x^{n}\end{array} \)
where ai‘s are real numbers such that an ≠0 for some natural number n.
As we know,
\(\begin{array}{l}\lim_{x\rightarrow a}x = a\end{array} \)
And
\(\begin{array}{l}\lim_{x\rightarrow a}x^2 = \lim_{x\rightarrow a}(x.x)=\lim_{x\rightarrow a}x.\lim_{x\rightarrow a}x=a.a=a^2\end{array} \)
Similarly, we can write the limit for xn as:
\(\begin{array}{l}\lim_{x\rightarrow a}x^n = a^n\end{array} \)
From these, we can write the limit of the polynomial function
\(\begin{array}{l}f(x)=a_{0}+a_{1}x+a_{2}x^{2}+….+a_{n}x^{n}\end{array} \)
by considering each term, i.e. a0, a1x, a2x2,…anxn are the functions.
So,
\(\begin{array}{l}\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}[a_{0}+a_{1}x+a_{2}x^{2}+….+a_{n}x^{n}]\end{array} \)
Split the limit for each term of the function as:
\(\begin{array}{l}\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}a_{0}+\lim_{x\rightarrow a}a_{1}x+\lim_{x\rightarrow a}a_{2}x^{2}+….+\lim_{x\rightarrow a}a_{n}x^{n}\end{array} \)
Now, take the coefficients out of the limit,
\(\begin{array}{l}\lim_{x\rightarrow a}f(x)=a_{0}+a_{1}\lim_{x\rightarrow a}x+a_{2}\lim_{x\rightarrow a}x^{2}+….+a_{n}\lim_{x\rightarrow a}x^{n}\end{array} \)
Therefore,
\(\begin{array}{l}\lim_{x\rightarrow a}f(x)=a_{0}+a_{1}a+a_{2}a^{2}+….+a_{n}a^{n} = f(a)\end{array} \)
\(\begin{array}{l}\large \lim_{x\rightarrow a}f(x) = f(a)\end{array} \) |
Let’s have a look at the example given below to understand how to apply limits for polynomial functions.
Example 1:
Compute:
\(\begin{array}{l}\lim_{x\rightarrow 2}[x^{3}-x^{2}+1]\end{array} \)
Solution:
\(\begin{array}{l}\lim_{x\rightarrow 2}[x^{3}-x^{2}+1]\end{array} \)
We know that,
\(\begin{array}{l}\lim_{x\rightarrow a}f(x) = f(a)\end{array} \)
Now,
\(\begin{array}{l}\lim_{x\rightarrow 2}[x^{3}-x^{2}+1]=(2)^3 – (2)^2+1\end{array} \)
= 8 – 4 + 1
= 5
Therefore,
\(\begin{array}{l}\lim_{x\rightarrow 2}[x^{3}-x^{2}+1]=5\end{array} \)
Limits of Rational Functions
Before applying limits for rational functions, first, understand what rational functions are and how to write them.
A function f is called a rational function, if
\(\begin{array}{l}f(x)=\frac{g(x)}{h(x)}\end{array} \)
, where g(x) and h(x) are polynomial functions such that h(x) ≠0.
The application of limit for f(x) as x tends to a is given as:
\(\begin{array}{l}\large \lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}\frac{g(x)}{h(x)} = \frac{\lim_{x\rightarrow a}g(x)}{\lim_{x\rightarrow a}h(x)}=\frac{g(a)}{h(a)}\end{array} \)
If h(a) = 0, then we consider two cases to define the limit of the rational function g(x)/h(x).
Case 1: g(a) ≠0
In this case, g(a) ≠0 and h(a) = 0 so the limit does not exist.
Case 2: g(a) = 0
Here, we need to write the functions g(x) and h(x) as:
g(x) = (x – a)k g1(x), where k is the maximum of powers of (x – a) in g(x)
h(x) = (x – a)l h1(x) as h (a) = 0
Assume that k > l, then we have
\(\begin{array}{l}\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}\frac{g(x)}{h(x)} = \frac{\lim_{x\rightarrow a}g(x)}{\lim_{x\rightarrow a}h(x)}=\frac{\lim_{x\rightarrow a}(x-a)^kg_{1}(x)}{lim_{x\rightarrow a}(x-a)^lh_{1}(x)}\end{array} \)
By applying the quotient rule of exponents,
\(\begin{array}{l}=\frac{\lim_{x\rightarrow a}(x-a)^{(k-l)}g_{1}(x)}{lim_{x\rightarrow a}h_{1}(x)}\end{array} \)
Now, by applying the limit,
\(\begin{array}{l}=\frac{0.g_{1}(a)}{h_{1}(a)}\end{array} \)
= 0
However, when k < l, the limit does not exist.
Thus, we need to evaluate the functions individually that are involved in the rational functions at the prescribed points at first. If this is of the form 0/0, then if possible, we have to rewrite the function such that we could cancel the factors which are causing the limit to be of the form 0/0.
Get more information about rational functions here.
Example 2:
Evaluate:
\(\begin{array}{l}\lim_{x\rightarrow 3}\frac{4x^2 -6x-18}{x^2 – 9}\end{array} \)
Solution:
\(\begin{array}{l}\begin{align*} \lim_{x\rightarrow 3}\frac{4x^2 -6x-18}{x^2 – 9}& =\frac{\lim_{x\rightarrow 3}(4x^2 -6x-18)}{\lim_{x\rightarrow 3}(x^2-9)}\\&=\frac{[4(3)^2-6(3)-18)]}{(3)^2-9}\\&=\frac{(36-18-18)}{9-9}\\&=\frac{0}{0}\end{align*}\end{array} \)
By evaluating the function at 2, we got 0/0.
So, let us factorise the functions to cancel the factors if possible.
4x2 – 6x – 18 = 4x2 – 12x + 6x – 18
= 4x(x – 3) + 6(x – 3)
= (x – 3)(4x + 6)
And x2 – 9 = x2 – 32 = (x – 3)(x + 3)
Thus,
\(\begin{array}{l}\begin{align*} \lim_{x\rightarrow 3}\frac{4x^2 -6x-18}{x^2 – 9}& =\frac{\lim_{x\rightarrow 3}(4x^2 -6x-18)}{\lim_{x\rightarrow 3}(x^2-9)}\\&=\frac{\lim_{x\rightarrow 3}(x-3)(4x+6)}{\lim_{x\rightarrow 3}(x+3)(x-3)}\\&=\frac{\lim_{x\rightarrow 3}(4x+6)}{\lim_{x\rightarrow 3}(x+3)}\\&=\frac{4(3)+6}{3+3}\\&=\frac{18}{6}\\&=3\end{align*}\end{array} \)
Example 3:
Compute:
\(\begin{array}{l}\lim_{x\rightarrow 2}\frac{x^2 – 100}{3x + 6}\end{array} \)
Solution:
\(\begin{array}{l}\lim_{x\rightarrow 2}\frac{x^2 – 100}{3x + 6} = \frac{\lim_{x\rightarrow 2}(x^2- 100)}{\lim_{x\rightarrow 2}(3x + 6)} = \frac{(2)^2 – 100}{3(2)+6}=\frac{4-100}{6+6}=\frac{-96}{12}=-8\end{array} \)
Similarly, we can compute the limits at specified points for a given rational function in different ways.
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