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 + . . . + anxn
♦ 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.
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 + . . . + anxn
• 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 fn(x) = anxn
• Thus we get:
f(x) = f0(x) + f1(x) + f2(x) + f3(x) + . . . + fn(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:
$$\lim_{x\rightarrow a} f(x) = f(a)$$
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.
In the next section, we will see some solved examples.
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