Question

lim┬(x→1)⁡〖(〖5x〗^2+3x+1)/(2x^2-4x-5)〗

Answer 1

$\lim _{x\to \:1}\left(\frac{5x^2+3x+1}{2x^2-4x-5}\right)$
$Sustituir\:la\:variable:$
$=\frac{5\cdot \:1^2+3\cdot \:1+1}{2\cdot \:1^2-4\cdot \:1-5}$
$=-\frac{9}{7}$ $\left(\mathrm{Decimal:\quad }\:-1.28571\dots \right)$