Question

compute limit x->3 x^4-81/2x^2-5x-3

Answer 1

\[(x^{4}-81)/(2x ^{2}-5x-3)\]

Answer 2

you asking for the limit of that is it goes to 3?

Answer 3

You have something called an 'indeterminate form' since both the numerator and denominator approach 0 as x approaches 3. In these cases, you can use L'Hopital's rule, which says\[\lim_{x \rightarrow a}\frac{f(x)}{g(x)}=\lim_{x \rightarrow a}\frac{f'(x)}{g'(x)}\]In your case, \[\lim_{x \rightarrow 3}\frac{x^4-81}{2x^2-5x-3}=\lim_{x \rightarrow 3}\frac{4x^3}{4x-5}=\frac{108}{7}\]