Question

Evaluate the Given limit: $\lim_{x\rightarrow 3}$ $\frac{x^4-81}{2x^2-5x-3}$

Answer 1

At x=2, the value of the given rational function takes the form 0/0 
=$\lim_{x\rightarrow 3}$ $\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$
=$\lim_{x\rightarrow 3}$ $\frac{(x+3)(x^2+9)}{2x+1}$
=$\frac{(3+3)(32+9)}{2(3)+1}$
=$\frac{6\times 18}{7}$
=$\frac{108}{7}$