# 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 $1$.

A best free mathematics education website for students, teachers and researchers.

###### Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

###### Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

###### Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.