0
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$$\lim_{x \to \infty} (\sqrt{x^2-49}-\sqrt{x^2-16} ) $$

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.

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

The denominator goes to $+\infty$

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1
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You forgot to divide with $\sqrt{x^2-49}+\sqrt{x^2-16}$

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