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> $$\lim\_{x \to \infty} (\sqrt{x^2-49}-\sqrt{x^2-16} ) $$ 
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I multiplied by the conjugate radical expression: 

$$=(\sqrt{x^2-49}-\sqrt{x^2-16}) \times (\sqrt{x^2-49}+\sqrt{x^2-16}) $$ 

$$= x^2-49-(x^2-16)=x^2-49-x^2+16=-33$$ 

$$\lim\_{x \to \infty}f(x) = -33$$ 

This is wrong. The correct answer is $0$. What is wrong in my process? Is it possible to solve this limit without multiplying by the conjugate? Thanks.