# Limit of $\dfrac{x^n-a^n}{x-a}$ when $x$ tends to $a$

## Formula

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

### Proof

$x$ and $a$ are two literals. Multiply $x$ by the same literal $n$ number of times and the product of them is denoted by $x^n$. Similarly, multiply $a$ by the same literal and same number of times and the product of them is represented by $a^n$ by the exponentiation.

The division of the subtraction of $a^n$ from $x^n$ by the subtraction of $a$ from $x$, formed an algebraic function in division form.

$\dfrac{x^n-a^n}{x-a}$

It is written in the following form when the value of the function is evaluated as the value of $x$ approaches $a$.

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

Now, let us try to find the value of the function.

01

#### Testing the functionality as x tends to a

Substitute $x = a$ to understand the functionality of the function as $x$ tends to $a$.

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

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

The function gets an indeterminate form. Hence, the function should be evaluated in the alternative method.

02

#### Transform the function

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

If $x \,\to\, a$, then $x-a \,\to\, 0$. So, change the value of the limit of the function.

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

Take $x-a = h$, then $x = a+h$. Now, eliminate the $x$ from the function.

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

Take $a^n$ common from both terms of the numerator.

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

03

#### Apply Binomial theorem

According to binomial theorem, the function ${\Bigg(1+\dfrac{h}{a}\Bigg)}^n$ can be expanded. So, expand the function ${\Bigg(1+\dfrac{h}{a}\Bigg)}^n$ to move ahead in the simplification.

$= \,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 0} \normalsize \dfrac{a^n \Bigg[\Bigg(1+\dfrac{n}{1!} \dfrac{h}{a} + \dfrac{n(n-1)}{2!} {\Bigg(\dfrac{h}{a}\Bigg)}^2 + \dfrac{n(n-1)(n-3)}{3!} {\Bigg(\dfrac{h}{a}\Bigg)}^3 + \cdots \Bigg) -1\Bigg]}{h}$

$= \,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 0} \normalsize \dfrac{a^n \Bigg[1+\dfrac{n}{1!} \dfrac{h}{a} + \dfrac{n(n-1)}{2!} {\Bigg(\dfrac{h}{a}\Bigg)}^2 + \dfrac{n(n-1)(n-3)}{3!} {\Bigg(\dfrac{h}{a}\Bigg)}^3 + \cdots -1\Bigg]}{h}$

$\require{cancel} = \,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 0} \normalsize \dfrac{a^n \Bigg[\cancel{1}+\dfrac{n}{1!} \dfrac{h}{a} + \dfrac{n(n-1)}{2!} {\Bigg(\dfrac{h}{a}\Bigg)}^2 + \dfrac{n(n-1)(n-3)}{3!} {\Bigg(\dfrac{h}{a}\Bigg)}^3 + \cdots -\cancel{1} \Bigg]}{h}$

$= \,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 0} \normalsize \dfrac{a^n \Bigg[\dfrac{n}{1!} \dfrac{h}{a} + \dfrac{n(n-1)}{2!} {\Bigg(\dfrac{h}{a}\Bigg)}^2 + \dfrac{n(n-1)(n-3)}{3!} {\Bigg(\dfrac{h}{a}\Bigg)}^3 + \cdots \Bigg]}{h}$

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

$= \,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 0} \normalsize \dfrac{a^n \Bigg[\dfrac{nh}{a} + \dfrac{n(n-1)h^2}{2a^2} + \dfrac{n(n-1)(n-3)h^3}{6a^3} + \cdots \Bigg]}{h}$

$h$ is a common multiplying factor in each term of the numerator and there is also one $h$ term in the denominator. So, take $h$ common from all the terms of the numerator.

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

$\require{cancel} = \,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 0} \normalsize \dfrac{a^n \times \cancel{h} \Bigg[\dfrac{n}{a} + \dfrac{n(n-1)h}{2a^2} + \dfrac{n(n-1)(n-3)h^2}{6a^3} + \cdots \Bigg]}{\cancel{h}}$

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

04

#### Evaluate the limit of the function

Substitute $h = 0$ to find the value of the limit of the function as $h$ approaches zero.

$= \,\,\,$ $a^n \Bigg[\dfrac{n}{a} + \dfrac{n(n-1)(0)}{2a^2} + \dfrac{n(n-1)(n-3){(0)}^2}{6a^3} + \cdots \Bigg]$

$= \,\,\,$ $a^n \Bigg[\dfrac{n}{a} + 0 + 0 + \cdots \Bigg]$

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

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

Use quotient rule of exponents to simplify the divisional function.

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

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

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

This is a standard result for a function which is in this form and it is used as a formula while dealing functions which are in this form in limits.

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