Question

Calcule o Limite

$\displaystyle \lim_{x \to p} \frac{x^n-p^n}{x-p} \, \,\, \, (n\ \textgreater \ 0 \, \in \, \mathbb{N})$

Answer 1

Considerando que:

$\displaystyle x^n-p^n=(x-p) \cdot (x^{n-1}+px^{n-2}+p^2x^{n-3}+ \, ... \, +p^{n-2}x+p^{n-1})$

Então temos:

$\displaystyle \lim_{x \to p} \frac{x^n-p^n}{x-p} \\ \\ \\ \lim_{x \to p} \frac{(x-p) \cdot (x^{n-1}+px^{n-2}+p^2x^{n-3}+ \, ... \, +p^{n-2}x+p^{n-1})}{(x-p)} \\ \\ \\ \lim_{x \to p} \, x^{n-1}+px^{n-2}+p^2x^{n-3}+ \, ... \, +p^{n-2}x+p^{n-1} \\ \\ \\ \lim_{x \to p} \, p^{n-1}+p \cdot p^{n-2}+p^2 \cdot p^{n-3}+ \, ... \, +p^{n-2}p+p^{n-1} \\ \\ \\ \lim_{x \to p} \, p^{n-1}+p^{n-1}+p^{n-1}+ \, ... \, + p^{n-1}+p^{n-1} \\ \\ \\ {\boxed{\boxed{\lim_{x \to p} f(x) = n \cdot p^{n-1}}}$

Answer 2

Olá!

$\\ \displaystyle \mathsf{\lim_{x \to p} \frac{x^n - p^n}{x - p} =} \\\\\\ \mathsf{\lim_{x \to p} \frac{(x - p) \cdot (x^{n - 1} + x^{n - 2} \cdot p^1 + x^{n - 3} \cdot p^2 + ... + x^1 \cdot p^{n - 2} + p^{n - 1})}{x - p} =} \\\\\\ \mathsf{\lim_{x \to p} x^{n - 1} + x^{n - 2} \cdot p^1 + x^{n - 3} \cdot p^2 + ... + x^1 \cdot p^{n - 2} + p^{n - 1} =} \\\\\\ \mathsf{p^{n - 1} + p^{n - 2} \cdot p + p^{n - 3} \cdot p^2 + p^{n - 4} \cdot p^3 + ... + p^1 \cdot p^{n - 2} + p^{n - 1} =} \\\\ \mathsf{p^{n - 1} + p^{n - 1} + p^{n - 1} + p^{n - 1} + ... + p^{n - 1} + p^{n - 1} + p^{n - 1} =} \\\\ \boxed{\mathsf{n \cdot p^{n - 1}}}$