Question

Find limit x to 5 (x^5 -3125)/ (x^3-125)​

Answer 1

Check the attachment.

Answer 2

The value of $\lim_{x \to 5} \dfrac{x^5-3125}{x^3-125}$ = $\dfrac{125}{3}$

Step-by-step explanation:

We have:

$\lim_{x \to 5} \dfrac{x^5-3125}{x^3-125}$

To find, the value of $\lim_{x \to 5} \dfrac{x^5-3125}{x^3-125}$ = ?

∴ $\lim_{x \to 5} \dfrac{x^5-3125}{x^3-125}$

Dividing numerator and denominator by (x - 5), we get

= $\lim_{x \to 5} \dfrac{\dfrac{x^5-5^5}{x-5}}{\dfrac{x^3-5^3}{x-5}}$

We know that,

The limit formula,

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

= $\dfrac{5(5)^{5-1} }{3(5)^{3-1}}$

= $\dfrac{5(5)^{4} }{3(5)^{2}}$

= $\dfrac{5}{3}\times (5)^{4-2}$

= $\dfrac{5}{3}\times (5)^{2}$

= $\dfrac{125}{3}$

Thus, the value of $\lim_{x \to 5} \dfrac{x^5-3125}{x^3-125}$ = $\dfrac{125}{3}$