<s>[INST] Calculate the limit at x=0 Find the limit of $f(x)=\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x}-\sqrt{a+x}}$ (at x=0) so that $f(x)$ becomes continuous for all $x$. My answer is $2\sqrt{a}$. Am I right?
Sorry to state the question incorrectly. We have to define $f(x)$ at $x=0$ such that $f(x)$ is continuous for all $x$. In that case my answer is $2\sqrt{a}$.
Let f(x)=$[x]$+$[-x]$ be a function where [.] stands for the greatest integer not greater than x. For any integer $m$, what can we say about $lim_{x \to m}$. Is $f(x)$ contiuous at $x=m$. Sorry for asking such vague question but I am forgetting the exact wording and the options given. This was a question asked in a class test.
 [/INST] The answer should be $\sqrt{a}$ assuming $a \geq 0$. Compute the limit of $f(x)$ as $x \to 0$. So for the function to be continuous at $x=0$, the functional value must be set equal to $\lim_{x \to 0} f(x)$.
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Note that $f(x)$ is not defined at $x=0$. Assuming $x \neq 0$. $$\begin{align} f(x) & = \dfrac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x} - \sqrt{a+x}}\\ & = \dfrac{(a^2-ax+x^2)-(a^2+ax+x^2)}{(a-x) - (a+x)} \times \dfrac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}\\ & = \dfrac{-2ax}{-2x} \times \dfrac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}\\ & = a \times \dfrac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} \end{align} $$ We will also assume that $a \geq 0$. Hence, $$\lim_{x \to 0} f(x) = \lim_{x \to 0} a \times \dfrac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} = a \times \dfrac{2 \sqrt{a}}{2a} = \sqrt{a}$$

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