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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.

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$f(x) = 0$? $ \ \ $ – TMM Jun 24 '12 at 18:56
A typo must surely exist in the numerator of this fraction. Do you mean for the middle terms under the radical to have different signs? – ncmathsadist Jun 24 '12 at 19:03
Oh ... i didn't see that. – Santosh Linkha Jun 24 '12 at 19:03
Echoing others, $f(x) = 0$. – copper.hat Jun 24 '12 at 19:06
@KunalSuri Each post must contain only one question. Kindly post your second question as a separate post. – user17762 Jun 24 '12 at 19:09

As $x\to 0$ you have $$\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}= {-2ax\over \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2}}\sim {-2ax\over 2a} = -x $$ and $${1\over \sqrt{a-x}-\sqrt{a+x}} = {-2x\over \sqrt{a-x}+\sqrt{a+x}}\sim -{x\over \sqrt{a}} $$ Now divide to see tht you get $\sqrt{a}$ for the limit of the ratio.

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EDIT:: $$ \frac{\sqrt{a^2}-\sqrt{a^2}}{\sqrt a - \sqrt a} = \frac{0}{0}\neq 2\sqrt a $$ You have to multiply by conjugate of both terms (in numerator and denominator) and get the following.

$$ \frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x}-\sqrt{a+x}} \\ = \frac{(a^2-ax+x^2)-(a^2+ax+x^2)}{(a-x)-(a+x)} \times \frac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} \\ = \frac{-2ax}{-2x}\times \frac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} \\ \text{ taking limit x } \rightarrow 0 \text{ we get} =a \frac{2\sqrt a}{2 a} = \sqrt a$$

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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 $\lim_{x \to 0} f(x)$.

Move your mouse over the gray area for the complete answer.

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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Great minds all thinking alike at once..... Cool! – ncmathsadist Jun 24 '12 at 19:10
This is my approach:- We have $\begin{align} f(x) & = \dfrac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x} - \sqrt{a+x}}\end{align}$ $\begin{align} lim_{x\to0}f(x) & = lim_{h\to0}\dfrac{\sqrt{a^2-ah+h^2}-\sqrt{a^2+ah+h^2}}{\sqrt{a-h} - \sqrt{a+h}}\end{align}$ $\begin{align}& = \dfrac{\sqrt{a^2-0+0}-\sqrt{a^2+0+0}}{\sqrt{a-0} - \sqrt{a+0}}\end{align}$ $\begin{align}& = \dfrac{\sqrt{a^2}-\sqrt{a^2}}{\sqrt{a} - \sqrt{a}}\end{align}$ . Here I carried out the usual algebraic manipulations and got $2\sqrt{a}$ – kusur Jun 24 '12 at 19:23

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