$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a} \,=\, n.a^{\displaystyle n-1}}$
Let $x$ be a variable, and $a$ and $n$ be two constants. Let’s assume that two quantities are expressed in exponential form as $x^{\displaystyle n}$ and $a^{\displaystyle n}$.
The ratio of the above two indeterminate quantities is written as follows in mathematics.
$\dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}$
The limit of this rational expression as the value of $x$ approaches to $a$ is written in the following mathematical form.
$\implies$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}}$
The limit of the $x$ raised to the power $n$ minus $a$ raised to the power $n$ by $x$ minus $a$ as the value of $x$ is closer to $a$, is equal to the $n$ times $a$ raised to the power $n$ minus $1$.
$\implies$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\displaystyle n}-a^{\displaystyle n}}{x-a}}$ $\,=\,$ $n \times a^{\displaystyle n-1}$
It can be called the power-difference limit rule in ratio form.
Learn how to derive the power-difference law of limits in ratio form in calculus mathematically.
List of the questions on power-difference property of limits in ratio form with solutions.
A best free mathematics education website for students, teachers and researchers.
Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.
Learn how to solve the maths problems in different methods with understandable steps.
Copyright © 2012 - 2022 Math Doubts, All Rights Reserved