Question

limx->-1 x^3-2x-1/x^5-2x-1 ? without using l'hospital rule

Answer 1

what do you get when you replace x by -1?

Answer 2

if you get
\[\frac{0}{0}\] then you can factor and cancel. it will be
\[\frac{(x+1)\times \text{something}}{(x+1)\times \text{something else}}\]

Answer 3

cancel and then replace x by -1.
in fact it will be
\[\frac{x^2-x-1}{x^4-x^3+x^2-x-1}\]

Answer 4

i got it :) thanks