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

1. anonymous

$(x^{4}-81)/(2x ^{2}-5x-3)$

2. anonymous

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

3. anonymous

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, $\lim_{x \rightarrow 3}\frac{x^4-81}{2x^2-5x-3}=\lim_{x \rightarrow 3}\frac{4x^3}{4x-5}=\frac{108}{7}$