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Posted December 12th, 2009 by alpha
in

$ \mathop{\lim}\limits_{x \to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}} $

‹ second derivative Trigonometric Limits ›

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~~THANK YOU ~~

On December 13th, 2009 alpha says:

$ \Gamma h\alpha\eta\kappa \Upsilon\Theta\mu $

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``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On December 12th, 2009 Structure says:

$ \mathop{\lim}\limits_{x \to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}} =\lim_{x\to a}\frac{a-x}{\sqrt{a+2x}+\sqrt{3x}}\frac{\sqrt{3a+x}+2\sqrt{x}}{3a-3x}=\frac{4\sqrt a}{6\sqrt {3a}}=\frac{2}{3 \sqrt 3} $

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