Question

To Prove :-

${ { \orange {\bf { \displaystyle \lim_{x \to a} \bigg( \dfrac{x^{n} - a^{n}}{x-a}\bigg) = n \cdot a^{n-1} }}}}$

At first , don't use direct formula _/\_

Use epsilon - delta definition , if you don't know it or had mistakes then only use direct formula _/\_ ​

Answer 1

Answer:

$In \: the \: given \: case \: f(x)= (x^n -a^n) ,g(x)=x-a$

$lim x->a (x^n -a^n)= lim x->a (x-a)=0 and g’(x)=1$

Hence the limit will be

$=lim x->a {(x^n -a^n) /(x-a)}</p><p>$

$= lim x->a {(nx^n-1-0) /1} </p><p></p><p>=na^(n-1)$

Answer 2

Step-by-step explanation:

Put $\sf\Delta x=x-a$ so that $\sf\Delta x \rightarrow 0$ as $\sf x \rightarrow a$ and $\sf\left|\frac{\Delta x}{a}\right|<1$

$\tt \red{ \leadsto\dfrac{x^{n}-d^{n}}{x-a}=\dfrac{(a+\Delta x)^{n}-d^{n}}{\Delta x}=\dfrac{d^{n}\left(1+\dfrac{\Delta x}{a}\right)^{n}-d^{n}}{\Delta x}}$

Applying Newton's Binomial Theorem for rational index we have,

$\blue{ \leadsto\left(1+\frac{\Delta x}{a}\right)^{n}=1+\left(\begin{array}{l}n \\ 1\end{array}\right)\left(\frac{\Delta x}{a}\right)+\left(\begin{array}{l}n \\ 2\end{array}\right)\left(\frac{\Delta x}{a}\right)^{2}+\left(\begin{array}{l}n \\ 3\end{array}\right)\left(\frac{\Delta x}{a}\right)^{3}+\ldots+\left(\begin{array}{l}n \\ r\end{array}\right)\left(\frac{\Delta x}{a}\right)^{r}+\ldots }$

$\purple{\sf{ \leadsto \qquad\therefore \quad\frac{x^{n}-a^{n}}{x-a} \: =\frac{a^{n}\left[1+\left(\begin{array}{l} \sf n \\ 1\end{array}\right)\left(\frac{\Delta x}{a}\right)+\left(\begin{array}{l} \sf n \\ 2\end{array}\right)\left(\frac{\Delta x}{a}\right)^{2}+\ldots+\left(\begin{array}{l} \sf n \\ \sf r\end{array}\right)\left(\frac{\Delta x}{a}\right)^{r}+\ldots\right]-a^{n}}{\Delta x}}}$

$\green{\tt{=\frac{\left[\left(\begin{array}{l} \tt n \\ 1\end{array}\right) a^{n-1} \Delta x+\left(\begin{array}{l} \tt n \\ 2\end{array}\right) a^{n-2}(\Delta x)^{2}+\ldots+\left(\begin{array}{l} \tt n \\ \tt r\end{array}\right) a^{n-r}(\Delta x)^{r}+\ldots\right]}{\Delta x}}}$

$\tt \orange{=\left(\begin{array}{l} \tt n \\ 1\end{array}\right) a^{n-1}+\left(\begin{array}{l} \tt n \\ 2\end{array}\right) a^{n-2}(\Delta x)+\ldots+\left(\begin{array}{l} \tt n \\ \tt r\end{array}\right) a^{n-r}(\Delta x)^{r-1}+\ldots}$

$\tt =\left(\begin{array}{l} \tt n \\ 1\end{array}\right) a^{n-1}+terms \: \: \: containing \: \: \Delta \: x \: \: and \: \: higher \: \: powers \: \: of \: \: \Delta \: x .$

Since $\Delta x=x-a, x \rightarrow a$ means $\Delta x \rightarrow 0$ and therefore $\begin{aligned} \lim _{x \rightarrow a} \frac{x^{n}-d^{n}}{x-a} &=\lim _{\Delta x \rightarrow 0}\left(\begin{array}{l} n \\ 1 \end{array}\right) d^{n-1}+\lim _{\Delta x \rightarrow 0} \\ &=\left(\begin{array}{l} n \\ 1 \end{array}\right) a^{n-1}+0+0+\ldots=n a^{n-1} \quad \text { since }\left(\begin{array}{l} n \\ 1 \end{array}\right)=n \end{aligned}$

As an illustration of this result, we have the following examples.