Proof for Power difference rule of Limits in Ratio form

Let $x$ be a variable, $a$ and $n$ be two constants. The subtraction of $a$ raised to the power $n$ from $x$ raised to the power $n$ represents an indeterminate quantity in algebraic form. Similarly, the subtraction of $a$ from $x$ represents another unknown quantity.

The limit of ratio of both quantities as $x$ approaches to $a$ is written in mathematics as follows.

$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}}$

This standard form is called the power difference limit rule in ratio form and it is used as a formula in calculus. So, let us learn how to prove the power difference rule of limits in ratio form.

Evaluate Limit by Direct substitution

First of all, let us test the algebraic function in rational form as $x$ approaches to $a$ by the direct substitution method.

$\implies$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}}$ $\,=\,$ $\dfrac{a^{\displaystyle n}-a^{\displaystyle n}}{a-a}$

$\,=\,\, \dfrac{0}{0}$

It is evaluated that the limit of the function is indeterminate as the value of $x$ approaches to $a$. Hence, it is recommendable to use another method instead of direct substitution method.

Remove complexity of function by Transformation

The quantities in exponential form are subtracted in numerator. In this case, the values of $x$, $a$ and $n$ are unknown. Hence, it is not possible to find the difference of the quantities in exponential notation.

$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}}$

Therefore, let us try to convert the function into another form for removing the complexity in the function.

If $x \,\to\, a$, then $x-a \,\to\, a-a$. Therefore, $x-a \,\to\, 0$

$\implies$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}}$ $\,=\,$ $\displaystyle \large \lim_{x-a \,\to\, 0}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}}$

Take $x-a = h$, then $x = a+h$.

Now, eliminate the $x$ from the whole function by substituting suitable values.

$\implies$ $\displaystyle \large \lim_{x-a \,\to\, 0}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(a+h)^{\displaystyle n}-a^{\displaystyle n}}{h}}$

Now, let’s concentrate on the following mathematical expression.

$\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(a+h)^{\displaystyle n}-a^{\displaystyle n}}{h}}$

In the numerator, $a$ raised to the power $n$ is a second term and it can be taken out from the first term as well.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(a\times 1+1 \times h)^{\displaystyle n}-a^{\displaystyle n}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(a\times 1+\dfrac{a}{a} \times h\Bigg)^{\displaystyle n}-a^{\displaystyle n}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(a\times 1+\dfrac{a \times h}{a}\Bigg)^{\displaystyle n}-a^{\displaystyle n}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(a\times 1+a \times \dfrac{h}{a}\Bigg)^{\displaystyle n}-a^{\displaystyle n}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(a\times \bigg(1+\dfrac{h}{a}\bigg)\Bigg)^{\displaystyle n}-a^{\displaystyle n}}{h}}$

The first term in the numerator can be written as the product of two factors by using the power of a product rule.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle n} \times \bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}-a^{\displaystyle n}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle n} \times \bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}-a^{\displaystyle n} \times 1}{h}}$

In both terms of the numerator, the $a$ raised to the power $n$ is a common factor. Hence, it can be taken out common from them.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle n} \times \Bigg(\bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}-1\Bigg)}{h}}$

Simplify the function by expansion

The function $\bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}$ can be expanded by the Binomial Theorem.

$\bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}$ $\,=\,$ $1$ $+$ $\dfrac{n}{1!} \dfrac{h}{a}$ $+$ $\dfrac{n(n-1)}{2!} \bigg(\dfrac{h}{a}\bigg)^2$ $+$ $\dfrac{n(n-1)(n-3)}{3!} \bigg(\dfrac{h}{a}\bigg)^3$ $+$ $\ldots$

$\implies$ $\bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}-1$ $\,=\,$ $\dfrac{n}{1!} \dfrac{h}{a}$ $+$ $\dfrac{n(n-1)}{2!} \bigg(\dfrac{h}{a}\bigg)^2$ $+$ $\dfrac{n(n-1)(n-3)}{3!} \bigg(\dfrac{h}{a}\bigg)^3$ $+$ $\ldots$

$\implies$ $\bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}-1$ $\,=\,$ $\dfrac{n}{1} \dfrac{h}{a}$ $+$ $\dfrac{n(n-1)}{2} \bigg(\dfrac{h}{a}\bigg)^2$ $+$ $\dfrac{n(n-1)(n-3)}{6} \bigg(\dfrac{h}{a}\bigg)^3$ $+$ $\ldots$

$\implies$ $\bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}-1$ $\,=\,$ $n \times \dfrac{h}{a}$ $+$ $\dfrac{n(n-1)}{2} \times \dfrac{h^2}{a^2}$ $+$ $\dfrac{n(n-1)(n-3)}{6} \times \dfrac{h^3}{a^3}$ $+$ $\ldots$

$\implies$ $\bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}-1$ $\,=\,$ $\dfrac{nh}{a}$ $+$ $\dfrac{n(n-1)h^2}{2a^2}$ $+$ $\dfrac{n(n-1)(n-3)h^3}{6a^3}$ $+$ $\ldots$

Now, substitute the expansion of the function in the limit of the algebraic function.

$\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle n} \Bigg(\bigg(1+\dfrac{h}{a}\bigg)^{\displaystyle n}-1\Bigg)}{h}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle n} \bigg(\dfrac{nh}{a} + \dfrac{n(n-1)h^2}{2a^2} + \dfrac{n(n-1)(n-3)h^3}{6a^3} + \ldots \bigg)}{h}}$

In the numerator, there is a common factor $h$ in each term. So, it can be taken out common from them.

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle n} \times h \bigg(\dfrac{n}{a} + \dfrac{n(n-1)h}{2a^2} + \dfrac{n(n-1)(n-3)h^2}{6a^3} + \cdots \bigg)}{h}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle n} \times \cancel{h} \bigg(\dfrac{n}{a} + \dfrac{n(n-1)h}{2a^2} + \dfrac{n(n-1)(n-3)h^2}{6a^3} + \cdots \bigg)}{\cancel{h}}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize a^{\displaystyle n} \bigg(\dfrac{n}{a} + \dfrac{n(n-1)h}{2a^2} + \dfrac{n(n-1)(n-3)h^2}{6a^3} + \cdots \bigg)}$

Find the Limit of the Algebraic function

It is time to evaluate the limit of the algebraic function as $x$ tends to zero by the direct substitution.

$= \,\,\,$ $a^{\displaystyle n} \bigg(\dfrac{n}{a} + \dfrac{n(n-1)(0)}{2a^2} + \dfrac{n(n-1)(n-3){(0)}^2}{6a^3} + \ldots \bigg)$

$= \,\,\,$ $a^{\displaystyle n} \bigg(\dfrac{n}{a} + 0 + 0 + \ldots \bigg)$

$= \,\,\,$ $a^{\displaystyle n} \times \dfrac{n}{a}$

$= \,\,\,$ $\dfrac{a^{\displaystyle n} \times n}{a}$

$= \,\,\,$ $\dfrac{a^{\displaystyle n}}{a} \times n$

$= \,\,\,$ $n \times \dfrac{a^{\displaystyle n}}{a}$

Use quotient rule of exponents to simplify the expression.

$= \,\,\,$ $n \times a^{\displaystyle n-1}$

$= \,\,\,$ $n.a^{\displaystyle n-1}$

$\,\,\, \therefore \,\,\,\,\,\, \displaystyle \large \lim_{x \,\to\, a} \large \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a} \,=\, n.a^{\displaystyle n-1}$

Therefore, it is proved that the limit of the subtraction of $a$ raised to the power $n$ from $x$ raised to the power $n$ by $x$ minus $a$ as $x$ approaches to $a$ is equal to $n$ times $a$ raised to the power $n$ minus $1$.