Math Doubts

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

$x$ is a variable. $a$ and $n$ are constant. The limit of algebraic function $\dfrac{x^n-a^n}{x-a}$ as $x$ approaches $a$ is often appeared in calculus. So, this standard result can be used as a formula in limits to find the limit of algebraic functions.

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

Now, let’s learn how to derive the proof to find the limit of this algebraic function as $x$ approaches $a$ in calculus.

Evaluate Limit by Direct Substitution method

Let’s find the limit of the algebraic function as $x$ approaches $a$ firstly.

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

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

As $x$ tends to $a$, the limit of this algebraic function is indeterminate. So, let’s try another method to find the limit of this algebraic function.

Transform Limit of Algebraic function

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

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

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

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

The $n$-th power of $a$ is common in both terms of the numerator. So, it can be taken common from both terms for simplifying the numerator further.

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

Simplify the Algebraic function

The function ${\Bigg(1+\dfrac{h}{a}\Bigg)}^n$ is same as the Binomial Theorem. So, it can be expanded by the Binomial Theorem.

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

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

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

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

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

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

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

$h$ 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.

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

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

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

Find the Limit of the Algebraic function

It is time to evaluate the limit of the algebraic function as x approaches zero.

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

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

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

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

Use quotient rule of exponents to simplify the expression.

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

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

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

Therefore, it is proved that the limit of the algebraic function as $x$ approaches $a$ is equal to $n$ times $a$ is raised to the power of $n-1$.



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