Use the epsilon definition of the limit that $\displaystyle \lim_{x\rightarrow \infty }f(x)=L$ if for every positive real number $\displaystyle \epsilon$, there exists a real number $\displaystyle M$ such that $\displaystyle \left | f(x)-L\right |< \epsilon $ whenever $\displaystyle x>M$, to prove $\displaystyle \lim_{x\rightarrow \infty }\frac{x+3}{x^2-3}=0$