becomes continuous for all x, is given by
A)
B)
C)
D)
Correct Answer: C
Solution :
\[f(x)=\frac{\sqrt{{{a}^{2}}-ax+{{x}^{2}}}-\sqrt{{{a}^{2}}+ax+{{x}^{2}}}}{\sqrt{a+x}-\sqrt{a-x}}\] f(x) is continuous for all x so it is continuous at \[\operatorname{x} = 0\] \[\therefore \,\, \,f(0) = \underset{x\,\to \,0}{\mathop{lim}}\, f(x)\] \[= \underset{x\,\to \,0}{\mathop{lim}}\,\frac{\sqrt{{{a}^{2}}-ax+{{x}^{2}}}-\,\,\sqrt{{{a}^{2}}+ax+{{x}^{2}}}}{\sqrt{a+x}-\sqrt{a-x}}\] \[= \underset{x\,\to \,0}{\mathop{lim}}\,\,\frac{-2ax}{2x}.\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \sqrt{a+x}+\sqrt{a-x} \right)}{\left( \sqrt{{{a}^{2}}-ax+{{x}^{2}}}\,+\,\sqrt{{{a}^{2}}+ax+{{x}^{2}}} \right)}\]\[=\,\,\frac{-a\times 2\sqrt{a}}{2a}=-\,\sqrt{a}\]
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