Question

compute limit[x->1] (x^3 -1)/(x^2 +5x -6)

Answer 1

aplly L HOPITAL RULE(differentiating numerator and denominator separately till the expression is coming out to be an in determinate form)

Answer 2

having the limit:
\[\lim_{x \rightarrow 1}\frac{ x ^{3}-1 }{ x ^{2}+5x-6 }\]
We can see it's a 0/0 indetermination, in a finite limit, we want to factor the numerator nd denominator in such way, that when we ecaluate the limit, we get no indetermination.
now, in the numerator we have a cube difference, and in the denominator 2nd degree equation, wich we can factor out:
\[\lim_{x \rightarrow 1}\frac{ (x-1)(x ^{2}+x+1) }{ (x-1)(x+6) }\]
We have (x-1) on numerator and denominator, we can wipe them out:
\[\lim_{x \rightarrow 1}\frac{ (x ^{2}+x+1) }{ (x+6) }\]
And evaluating the limit:
\[\lim_{x \rightarrow 1}\frac{ x ^{2}+x+1 }{ x+6 }=\frac{ 3 }{ 7 }\]

Answer 3

THANKS A LOT