Question

When limits tends to 1
2x^8-3x^2+1/x^8+6x^5-7

Answer 1

$\lim_{x\to1}\frac{2x^8-3x^2+1}{x^8+6x^5-7}$  $=\frac{5}{17}$

Step-by-step explanation:

$\lim_{x\to1}\frac{2x^8-3x^2+1}{x^8+6x^5-7}$       $[\frac{0}{0} form]$    applying L' hospital's rule  

$=\lim_{x\to1}\frac{16x^7-6x}{8x^7+30x^4}$

$=\frac{16-6}{8+30}$

$=\frac{10}{38}$

$=\frac{5}{17}$