Question

Prove that:

$\boxed{\lim \limits_{x \to a} \dfrac{x^n - a^n}{x-a} = n x^{n-1} } \ \textless \ br /\ \textgreater$
Without using L'Hôpital's rule.​

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Answer 1

$\huge\text{\underline{\underline{Question}}}$

*Check the typo.

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

$\huge\text{\underline{\underline{Explanation}}}$

Substituting $x=a$ to $x^{n}-a^{n}$ gives $a^{n}-a^{n}=0$.

$\Large\text{\{Factor theorem\}}$

$\large\text{ \boxed{\text{Explanation}} }$

The factor theorem states that a polynomial $f(x)$ is divisible by $(x-a)$ if $f(a)=0$. The converse also holds.

$x^{n}-a^{n}$ is divisible by $(x-a)$ from factor theorem.

It factorizes to, -

$\text{ \cdots\longrightarrow\boxed{x^{n}-a^{n}=(x-a)Q(x)} }$

$\text{ (Q(x) is the quotient of the division.)}$

By actual division from synthetic division, we can observe that -$\text{ \cdots\longrightarrow\boxed{Q(x)=x^{n-1}+ax^{n-2}+a^{2}x^{n-3}+\cdots+a^{n-1}} }$

The proof for the equation is -

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}&xQ(x)=x^{n}+&ax^{n-1}+&a^{2}x^{n-2}+&\cdots+&a^{n-1}x\\&aQ(x)=&ax^{n-1}+&a^{2}x^{n-2}+&\cdots+&a^{n-1}x+a^{n}\end{aligned}} }$

$\text{ \cdots\longrightarrow\boxed{(x-a)Q(x)=x^{n}-a^{n}} }$

Hence we proved that, -

$\text{ \cdots\longrightarrow\boxed{Q(x)=x^{n-1}+ax^{n-2}+a^{2}x^{n-3}+\cdots+a^{n-1}} }$

$\Large\text{\{Limits\}}$

$\large\text{ \boxed{\text{Explanation}} }$

The limiting value is the specified value of the function that approaches to, as variable approaches one value.

Let us find the limiting value.

Given limit

$\text{ =\displaystyle\lim_{x\to a}\dfrac{x^{n}-a^{n}}{x-a} }$

$=\displaystyle\lim_{x\to a}\dfrac{(x-a)Q(x)}{x-a}$

$=\displaystyle\lim_{x\to a}Q(x)$

$=\displaystyle\lim_{x\to a}\left(x^{n-1}+ax^{n-2}+a^{2}x^{n-3}+\cdots+a^{n-1}\right)$

$=na^{n-1}$

$\huge\text{\underline{\underline{Deeper guide}}}$

This fact is used in derivatives also. Let us prove the derivatives of polynomials.

$\bullet\ \displaystyle\lim_{x\to a}\left\{f(x)+g(x)\right\}=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$

$\bullet\ \displaystyle\lim_{x\to a}kf(x)=k\lim_{x\to a}f(x)$

$\text{ \bullet\ \displaystyle\lim_{x\to a}\dfrac{x^{n}-a^{n}}{x-a}=na^{n-1} }$

Let the polynomial be -

$\cdots\longrightarrow P(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n}x^{0}$

The derivative $P'(x)$ at $x=a$ is $P'(a)$.

$P'(a)$

$\displaystyle=\lim_{x\to a}\dfrac{P(x)-P(a)}{x-a}$

$\text{ =\displaystyle\lim_{x\to a}\left(a_{0}\cdot\dfrac{x^{n}-a^{n}}{x-a}+a_{1}\cdot\dfrac{x^{n-1}-a^{n-1}}{x-a}+a_{2}\cdot\dfrac{x^{n-2}-a^{n-2}}{x-a}+\cdots+a_{n}\cdot\dfrac{x^{0}-a^{0}}{x-a}\right) }$

$\text{ =\displaystyle\lim_{x\to a}\left(a_{0}\cdot\dfrac{x^{n}-a^{n}}{x-a}\right)+\cdots+\lim_{x\to a}\left(a_{n}\cdot\dfrac{x^{0}-a^{0}}{x-a}\right) }$

$\text{ =\displaystyle a_{0}\lim_{x\to a}\left(\dfrac{x^{n}-a^{n}}{x-a}\right)+\cdots+a_{n}\lim_{x\to a}\left(\cdot\dfrac{x^{0}-a^{0}}{x-a}\right) }$

$=\displaystyle a_{0}\cdot na^{n-1}+a_{1}\cdot (n-1)a^{n-2}+a_{2}\cdot (n-2)a^{n-3}+\cdots+a_{n}\cdot0$

$=(a_{0}x^{n})'+(a_{1}x^{n-1})'+(a_{2}x^{n-2})'+\cdots+(a_{n}x^{0})'\text{ at }x=a$

Hence we proved the derivatives of polynomials.