Chapter 13.5 - Limits of Polynomial and Rational Functions

In the previous section, we completed the examples on algebra of limits. In this section, we will see limits of polynomial functions and rational functions.

• We have learned about polynomial functions in a previous chapter (details here). Let us recall some important points. It can be written in 5 steps:
1. The general form of a polynomial function is:
f(x) = a₀ + a₁x + a₂x² + a₃x³ + . . . + a_nxⁿ
    ♦ Where a₀, a₁, a₂ . . . , are real numbers.
    ♦ Also, n should be a natural number.
2. We have already seen many polynomial functions in this chapter. Let us see some examples:
(i) f(x) = x² + 1 is a polynomial function.
• It can be rearranged as: f(x) = 1 + x²
    ♦ This is a polynomial function with two terms.
    ♦ Comparing with the general form in (1), we get:
    ♦ a₀ = 1, a₁ = 0 and a₂ = 1  
(ii) f(x) = x is a polynomial function.
    ♦ This is a polynomial function with one term.
    ♦ Comparing with the general form in (1), we get:
    ♦ a₀ = 0 and a₁ = 1
(iii) f(x) = x³ + 2x² + 3x + 5 is a polynomial function.
• It can be rearranged as: f(x) = 5 + 3x + 2x² + x³
    ♦ This is a polynomial function with four terms.
    ♦ Comparing with the general form in (1), we get:
    ♦ a₀ = 5, a₁ = 3, a₂ = 2 and a₃ = 1
3. Consider the constant function f(x) = $\lambda$
    ♦ We know that $\lambda$ can be any real number.
    ♦ Suppose that $\lambda$ = 4.
• Then we get: f(x) = 4
• This can be written as: f(x) = 4x⁰ (∵ x⁰ is 1)
• f(x) = 4x⁰ is a polynomial function with one term.
    ♦ Comparing with the general form in (1), we get: a₀ = 4.
4. Consider again the constant function f(x) = $\lambda$
• If $\lambda$ is zero, then that constant function is called a zero function.
• For a zero function, whatever be the input x-value, the output value will be always zero.
• So the zero function can be written as f(x) = 0
• This is a polynomial function with one term.
    ♦ Comparing with the general form in (1), we get: a₀ = 0
5. So we can say that, most of the functions that we saw so far in this chapter are polynomial functions.
• However, we must remember that functions like $f(x) = x^{\frac{3}{2}} + 1$ are not polynomial functions. This is because, $\frac{3}{2}$ is not an integer.

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Our next task is to find a general form for the limit of polynomial functions. It can be written in 9 steps:
1. We know that, $\lim_{x\rightarrow a} x = a$
2. Next let us find $\lim_{x\rightarrow a} x^2$
• The algebra of limits respect multiplication. So we can write:
$\lim_{x\rightarrow a} x^2 = \lim_{x\rightarrow a} x \,. \,\lim_{x\rightarrow a} x = a\, .\, a = a^2$
3. Next let us find $\lim_{x\rightarrow a} x^3$
• The algebra of limits respect multiplication. So we can write:
$\lim_{x\rightarrow a} x^3 = \lim_{x\rightarrow a} x^2 \,. \,\lim_{x\rightarrow a} x = a^2\, .\, a = a^3$
4. We have the limit for x¹, x², and x³.
• We want the limit for xⁿ. For that, we can use mathematical induction. We have learned the principles of mathematical induction in a previous chapter (details here).
• For the present case, we can use five simple steps:
(i) We want to prove the statement: $\lim_{x\rightarrow a} x^n = a^n$
(ii) First we must prove the statement for n = 1: $\lim_{x\rightarrow a} x^1 = a^1$
• We already know that, this is true.
(iii) Next we assume that the statement is true for n = k
$\lim_{x\rightarrow a} x^k = a^k$
(iv) Next we must prove that, the statement is true for n = k+1
• That is., we must prove that: $\lim_{x\rightarrow a} x^{k+1} = a^{k+1}$
(v) This can be done as follows:

$\begin{array}{ll} {}&{\lim_{x\rightarrow a} x^{k+1}} & {~=~}& {a^{k+1}} &{} \\ {\Rightarrow}&{\lim_{x\rightarrow a} \left[x^k \, . \,x \right]} & {~=~}& {a^k \, .\, a} &{} \\ {\Rightarrow}&{\lim_{x\rightarrow a} x^k \, . \, \lim_{x\rightarrow a}x} & {~=~}& {a^k \, .\, a ~~ \color {green} {\text{- - - (A)}}} &{} \\ {\Rightarrow}&{a^k \, . \, a} & {~=~}& {a^k \, .\, a ~~ \color {green} {\text{- - - (B)}}} &{} \\ \end{array}$

◼ Remarks:
• We are able to write the line marked as A because, algebra of limits respects multiplication.
• We are able to write the line marked as B, using step (iii).

◼ Thus we have proved the statement in (iv). And so, the statement in (i) is proved.

5. Now consider the general form of a polynomial function:
f(x) = a₀ + a₁x + a₂x² + a₃x³ + . . . + a_nxⁿ
• We see that, there are several terms.
• We know that, functions can be added. So let each term be a separate function.
• We can write:
    ♦ Let f₀(x) = a₀
    ♦ Let f₁(x) = a₁x
    ♦ Let f₂(x) = a₂x²
    ♦ Let f₃(x) = a₃x³
    ♦ . . .
    ♦ Let f_n(x) = a_nxⁿ
• Thus we get:
f(x) = f₀(x) + f₁(x) + f₂(x) + f₃(x) + . . . + f_n(x)
6. Now we can write the limit of f(x):

$$\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} f_0(x) + \lim_{x\rightarrow a} f_1(x) + \lim_{x\rightarrow a} f_2(x) + \lim_{x\rightarrow a} f_3(x) + .~.~.~+\lim_{x\rightarrow a} f_n(x)$$
• We are able to write this because, algebra of limits respects addition.
7. Substituting the actual functions in (6), we get:
$$\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} a_0 + \lim_{x\rightarrow a} a_1 x + \lim_{x\rightarrow a} a_2 x^2 + \lim_{x\rightarrow a} a_3 x^3 + .~.~.~+\lim_{x\rightarrow a} a_n x^n$$
8. But from (4), we have: $\lim_{x\rightarrow a} x^n = a^n$
• So the result in (7) becomes:
$$\lim_{x\rightarrow a} f(x) = a_0 + a_1 a + a_2 a^2 + a_3 a^3 + .~.~.~+ a_n a^n$$
9. Consider the RHS of the result in (8). This RHS is f(a).
• So we can write:
$$\lim_{x\rightarrow a} f(x) = f(a)$$

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Now we will see rational functions. It can be written in 5 steps:
1. A function $f(x)$ is said to be a rational function if the following three conditions are satisfied:
(i) $f(x)$ can be written in the form $\frac{g(x)}{h(x)}$
(ii) $g(x)$ and $h(x)$ are polynomial functions.
(iii) $h(x)$ is not a zero function.
2. If f(x) is a rational function, then it’s limit as x approaches ‘a’ can be written as:

$$\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a}\left[\frac{g(x)}{h(x)} \right] = \frac{\lim_{x\rightarrow a} g(x)}{\lim_{x\rightarrow a} h(x)} = \frac{g(a)}{h(a)}$$
3. So we can write:
$$\lim_{x\rightarrow a} f(x) = \frac{g(a)}{h(a)}$$
4. Consider the result in (3)
• If the denominator h(a) is zero, we say that, the limit does not exist.
• This is because, division by zero will give a number which does not exist.
5. Consider again the result in (3)
• If both denominator h(a) and numerator g(a) are zero, we can rewrite the functions and cancel the factors which causes the result to be $\frac{0}{0}$
• This will become clear when we see some solved examples.

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In the next section, we will see some solved examples. 

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