How to calculate this limit at infinity?

Question

How would you calculate this limit:

$$ \lim_{a\to ∞} \left(\sqrt {x^2+a}-\sqrt {a}\right) $$

The answer is 0 but I have no idea how to get to it.

Answer 1

Hint $$\lim_{a\to ∞} \left(\sqrt {x^2+a}-\sqrt {a}\right)=\lim_{a\to ∞} \left(\sqrt {x^2+a}-\sqrt {a}\right)\bigg(\frac{\sqrt{x^2+a}+\sqrt{a}}{\sqrt{x^2+a}+\sqrt{a}}\bigg)$$

$$=\lim_{ a\rightarrow \infty} \frac{x^2}{\sqrt{x^2+a}+\sqrt{a}}$$

Answer 2

Hint: Try multiply with $$1=\frac{ \sqrt{x^2+a}+\sqrt{a} }{ \sqrt{x^2+a}+\sqrt{a} }$$

Answer 3

Multiply by $\sqrt{x^2+a} + \sqrt{a}\over {\sqrt{x^2+a} + \sqrt{a}}$, then use $(a+b)(a-b) = a^2-b^2$:

$$ \lim_{a\to ∞} \left(\sqrt {x^2+a}-\sqrt {a}\right) = \lim_{a\to ∞} \left((\sqrt {x^2+a}-\sqrt {a}) ( \sqrt{x^2+a} + \sqrt{a})\over {\sqrt{x^2+a} + \sqrt{a}}\right) = \lim_{a\to ∞} \left(x^2 \over {\sqrt{x^2+a} + \sqrt{a}}\right) $$

Answer 4

Maybe the physicist in me likes this approach $$ \sqrt{x^2+a}-\sqrt{a} =\sqrt{a}\sqrt{1+\frac{x^2}{a}}-\sqrt{a} $$ then we can expand $$ \sqrt{1+\frac{x^2}{a}}\to 1+\frac{x^2}{2a} +O(a^{-2})\approx 1+\frac{x^2}{2a} $$ then we have $$ \sqrt{a}+\frac{x^2}{2\sqrt{a}}+O(a^{-3/2})-\sqrt{a} $$ so in the limit $\frac{x^2}{a}<<1$ then we have $$ \sqrt{x^2+a}-\sqrt{a}\approx \frac{x^2}{2\sqrt{a}} $$ now take $a\to\infty$.

Or rationalize the numerator as per the other quicker/more beautiful answers show.