# Help with limit of radical expression

$$\lim_{x \to \infty} (\sqrt{x^2-49}-\sqrt{x^2-16} )$$

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.

• $(\sqrt{x^2-49}-\sqrt{x^2-16}) \times (\sqrt{x^2-49}+\sqrt{x^2-16})=-33$ – hhsaffar Nov 12 '13 at 23:32
• @hhsaffar thanks, I edited it to say $-33$. I still don't understand how the answer is $0$ though. – Emi Matro Nov 12 '13 at 23:36

$\lim_{x \to \infty} (\sqrt{x^2-49}-\sqrt{x^2-16} )=\lim_{x \to \infty}\frac{(\sqrt{x^2-49}-\sqrt{x^2-16} )(\sqrt{x^2-49}+\sqrt{x^2-16} )}{(\sqrt{x^2-49}+\sqrt{x^2-16} )}=\lim_{x \to \infty}\frac{-33}{(\sqrt{x^2-49}+\sqrt{x^2-16} )}=0$
The denominator goes to $+\infty$
You forgot to divide with $\sqrt{x^2-49}+\sqrt{x^2-16}$