Question

evaluate lim x tends to a ( √a+2x-√3x/√3a+x-2√x)​

Answer 1

EXPLANATION.

$\sf : \implies \: \lim_{x \to \: a} \: = \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} } \: \: \: \: and \: \:( a \: \ne \: 0) \\ \\ \sf : \implies \: we \: can \: put \: x \: = a \: in \: equation \: \\ \\ \sf : \implies \: \lim_{x \to \: a} \: = \frac{ \sqrt{a + 2a} \: - \sqrt{3a} }{ \sqrt{3a + a} - 2 \sqrt{a} } \\ \\ \sf : \implies \: \lim_{x \: \to \: a} \: = \frac{ \sqrt{3a} - \sqrt{3a} }{ \sqrt{4a } - 2 \sqrt{a} } \\ \\ \sf : \implies \: it \: is \: in \: \frac{0}{0} \: \: form \: indeterminant$

$\sf : \implies \: in \: the \: case \: of \: roots \: we \: can \: simply \: rationalize \: \\ \\ \sf : \implies \: \lim_{x \to \: a} \: = \frac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} } \: \: \: \times \: \: \: \frac{ \sqrt{a + 2x} + \sqrt{3x} }{ \sqrt{a + 2x} + \sqrt{3x} } \: \: \: \times \: \: \: \: \frac{ \sqrt{3a + x} \: + 2 \sqrt{x} }{ \sqrt{3a + x} \: + \: 2 \sqrt{x} } \\ \\ \sf : \implies \: \lim_{x \to \: a} \: = \frac{a + 2x - 3x( \sqrt{3a + x} + 2 \sqrt{x} ) }{3a + x - 4x( \sqrt{a + 2x} + \sqrt{3x}) } \\ \\ \sf : \implies \: \lim_{x \to \: a} \: = \frac{(a - x)( \sqrt{3a + x} + 2 \sqrt{x}) }{3(a - x)( \sqrt{a + 2x} + \sqrt{3x}) }$

$\sf : \implies \: \lim_{x \to \: a} \: = \dfrac{ \sqrt{3a + x} + 2 \sqrt{x} )}{3( \sqrt{a + 2x} + \sqrt{3x} ) } \\ \\ \sf : \implies \: put \: the \: value \: of \: x \: = a \: in \: equation \\ \\ \sf : \implies \: \lim_{x \to \: a} \: = \frac{ (\sqrt{4a} + 2 \sqrt{a} )}{3( \sqrt{3a} + \sqrt{3a}) } \\ \\ \sf : \implies \: \lim_{x \: \to \: a} \: = \frac{4 \sqrt{a} }{3 \times 2 \times \sqrt{3a} } \\ \\ \sf : \implies \: \lim_{x \: \to \: a} \: = \frac{2}{3 \sqrt{3} } \: = \: answer$