<s>[INST] How to prove the following in general: $\lim\limits_{x\rightarrow a} \frac{x^n-a^n}{x-a}=na^{n-1}$? To prove
$$\lim_{x\rightarrow a} \frac{x^n-a^n}{x-a}=na^{n-1}$$
The proof is easy when we take $n$ as positive integer and $a$ any positive real number.
In my book it is given that the result is true even when $n$ is any rational number and $a$ any positive real number. But the proof is not given.
Please provide some hint to construct the proof for general case.
 [/INST] $$\begin{array}{rcl}
\displaystyle \lim_{x\rightarrow a} \frac{x^n-a^n}{x-a}
&=& \displaystyle \lim_{x\rightarrow a} \frac{x^{p/q}-a^{p/q}}{x-a} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{\left(x^{p/q}-a^{p/q}\right)\left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{x^p - a^p}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{\left(x-a\right) \left(x^{p-1} + x^{p-2}a + \cdots + xa^{p-2} + a^{p-1}\right)}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{x^{p-1} + x^{p-2}a + \cdots + xa^{p-2} + a^{p-1}}{x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}} \\
&=& \displaystyle \frac{a^{p-1} + a^{p-2}a + \cdots + aa^{p-2} + a^{p-1}}{a^{(q-1)p/q} + a^{(q-2)p/q}a^{p/q} + \cdots + a^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}} \\
&=& \displaystyle \frac{pa^{p-1}}{qa^{(q-1)p/q}} \\
&=& \displaystyle \frac p q a^{(p-1)-(q-1)p/q} \\
&=& \displaystyle \frac p q a^{p-1-p+p/q} \\
&=& \displaystyle \frac p q a^{p/q-1} \\
&=& \displaystyle n a^{n-1} \\
\end{array}$$
</s>