Showing that $\sqrt{x^2 + a^2} - x \rightarrow 0$ as $x \rightarrow \infty$

Question

Hey guys I'm really struggling to prove this, any help would be appreciated!!

"Show that $\sqrt{x^2 + a^2} - x \rightarrow 0$ as $x \rightarrow \infty$. (Make use of $u - v = \frac{u^2 - v^2}{u + v})$."

Intuitively I understand as it basically becomes infinity - infinity once $x$ becomes sufficiently large, however I cannot seem to write a justifiable proof.

Answer 1

Hint

\begin{align}\sqrt{x^2+a^2}-x & \\& =\dfrac{(\sqrt{x^2+a^2}-x)(\sqrt{x^2+a^2}+x)}{\sqrt{x^2+a^2}+x}\\& =\dfrac{x^2+a^2-x^2}{\sqrt{x^2+a^2}+x}\\& =\dfrac{a^2}{\sqrt{x^2+a^2}+x}.\end{align}

Answer 2

We want to prove $$\lim_{x\rightarrow\infty} \sqrt{x^2+a^2}-x=0$$

To do this, we use $$\sqrt{x^2+a^2}-x=\frac{x^2+a^2-x^2}{\sqrt{x^2+a^2}+x}=\frac{a^2}{\sqrt{x^2+a^2}+x}$$

Now, the limit is obvious.

Answer 3

HINT

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Let $u=\sqrt {x^2+a^2} $ and $v=x $.