Math Doubts

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

$x$ is a variable and $a$ and $n$ are constants. They are used to form a complex function $\dfrac{x^n-a^n}{x-a}$.

The limit of $\dfrac{x^n-a^n}{x-a}$ as $x$ approaches $a$ is expressed mathematically in the following form to represent the value of function $\dfrac{x^n-a^n}{x-a}$ when the value of $x$ tends to $a$.

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

This type of functions often appears in calculus. So, it is used as a limit rule. Therefore, learn how to derive the proof for this function as $x$ approaches $a$ in calculus.

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 as $x$ approaches $a$. Hence, the function should be evaluated in the alternative method.

Transform the function

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.

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

Apply Binomial theorem

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

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

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

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

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

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

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

$h$ 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.

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

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

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

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 expression.

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

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

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

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.

Follow us
Email subscription
Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more