You will discover why if you factorise top and bottom.
x^4-81=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9).
2x^2-5x-3=(x-3)(2x+1)
There's a common factor x-3, so we remove it from top and bottom: (x+3)(x^2+9)/(2x+1). This factor is called a removable factor which becomes zero when we put x=3. That's why we get 0/0 when we try to evaluate the limit. When the factor is removed we can safely put x=3: (x+3)(x^2+9)/(2x+1)=6*18/7=108/7.