Question

Prove that:
$\boxed{\lim \limits_{x \to a} \dfrac{x^n - a^n}{x-a} = n x^{n-1} }$
Without using L'Hôpital's rule.​

Answer 1

$\Large\text{\underline{\underline{Question}}}$

Prove that $\displaystyle\lim_{x\to a}\dfrac{x^{n}-a^{n}}{x-a}=na^{n-1}$ without using L'Hôpital's rule.

$\Large\text{\underline{\underline{Explanation}}}$

Consider -

$\cdots\longrightarrow f(x)=x^{n}-a^{n}.$

By factor theorem, -

$\cdots\longrightarrow f(a)=0\iff f(x)=(x-a)Q(x).$

Assertion: -

$\text{ \cdots\longrightarrow f(x)=(x-a)(x^{n-1}+ax^{n-2}+\cdots+a^{n-2}x+a^{n-1}). \cdots[1] }$

Let us suppose $\text{[1]}$ for the base case $n=2$. Then, -

$\cdots\longrightarrow x^{2}-a^{2}=(x-a)(x+a).$

Let us suppose $n=k+1$. Then, -

$\cdots\longrightarrow x^{k+1}-a^{k+1}=(x-a)(x^{k}+ax^{k-1}+\cdots+a^{k-1}x+a^{k}).$

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}x^{k+1}+&ak^{k}+\cdots+a^{k}x&=x(x^{k}+ax^{k-1}+\cdots+a^{k-1}x+a^{k})\\\\&ax^{k}+\cdots+a^{k}x+a^{k+1}&=a(x^{k}+ax^{k-1}+\cdots+a^{k-1}x+a^{k})\end{aligned}} }$

We now subtracted and obtained -

$\cdots\longrightarrow x^{k+1}-a^{k+1}=(x-a)(x^{k}+ax^{k-1}+\cdots+a^{k-1}x+a^{k}).$

Hence, the factorization is true for all natural numbers $n\geq2$ by mathematical induction.

By the property of limits, -

$\text{ \cdots\longrightarrow\boxed{\displaystyle\begin{aligned}&\lim_{x\to a}\dfrac{x^{n}-a^{n}}{x-a}\\\\&=\lim_{x\to a}(x^{n-1}+ax^{n-2}+\cdots+a^{n-2}x+a^{n-1})\\\\&=a^{n-1}+a^{n-1}+\cdots+a^{n-1}+a^{n-1}\\\\&=na^{n-1}.\blacksquare\end{aligned}} }$

$\Large\text{\underline{\underline{More information}}}$

Let us recall that -

$\text{ \cdots\longrightarrow\boxed{\displaystyle\lim_{x\to a}\dfrac{x^{n}-a^{n}}{x-a}=na^{n-1}.} }$

Let us find $(x^{n})'$. Then, -

$\text{ \cdots\longrightarrow\displaystyle\boxed{\begin{aligned}&\lim_{h\to 0}\dfrac{(x+h)^{n}-x^{n}}{(x+h)-x}\\\\&=\lim_{h\to 0}\dfrac{(x+h)^{n-1}+(x+h)^{n-2}x+\cdots+(x+h)x^{n-2}+x^{n-1}}{h}\\\\&=x^{n-1}+x^{n-1}+\cdots+x^{n-1}+x^{n-1}\\\\&=nx^{n-1}.\end{aligned}} }$

Hence, we proved that the derivative of $x^{n}$ is equal to $nx^{n-1}$.