Question

$\footnotesize\sf \: If \: \lim\limits_{x\to a} \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} } \: \: equals \: k, \: then \: k \: is \: equals \: to :-$

$\sf \: a.) \dfrac{2}{3 \sqrt{3} }$

$\sf \: b.) \dfrac{2}{3}$

$\sf \: c.) \dfrac{3}{2}$

$\sf \: d.) \sqrt{3}$

Need complete solution and explanation, incorrect answer or answer without explanation will be reported immediately.
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Answer 1

⠀⠀⠀⌬⠀ Question :

$\sf \: If \: \lim\limits_{x\to a} \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} } \: \: equals \: k, \: then \: k \: is \: equals \: to :-$

$\sf \: a.) \dfrac{2}{3 \sqrt{3} }$

$\sf \: b.) \dfrac{2}{3}$

$\sf \: c.) \dfrac{3}{2}$

$\sf \: d.) \sqrt{3}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\twoheadrightarrow \sf \: \: \lim\limits_{x\to a} \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} } \: \: = \: k \:$

$\qquad \bigstar \underline {\pmb{\purple{\sf By \:Using \:the \:Given \: \::\:}}}$

$\twoheadrightarrow \sf \: \: \lim\limits_{x\to a} \dfrac{\sqrt{ a + 2x } - \sqrt{3x} }{\sqrt{3a + x } - 2\sqrt{x}} \: \times \: \dfrac{ \sqrt{a + 2x} + \sqrt{3x} }{ \sqrt{a + 2x} + \sqrt{3x} } \: \: = \: k \:$

$\twoheadrightarrow \sf \: \: \lim\limits_{x\to a} \dfrac{\sqrt{ a + 2x } - \sqrt{3x} }{\sqrt{3a + x } - 2\sqrt{x}} \: \times \: \dfrac{ \sqrt{a + 2x} + \sqrt{3x} }{ \sqrt{a + 2x} + \sqrt{3x} } \: \: = \: k \:$

$\twoheadrightarrow \sf \: \: \lim\limits_{x\to a} \dfrac{ a + 2x - 3x }{\big( \sqrt{3a + x } - 2\sqrt{x}\big) \big( \sqrt{a + 2x} + \sqrt{3x} \big) } \:\: \: = \: k \:$

$\twoheadrightarrow \sf \: \: \lim\limits_{x\to a} \dfrac{ \big( a - x \big) \big[ \sqrt{3a + x } + 2\sqrt{x} \big] }{\big( 3a + x - 4x \big) \big( \sqrt{a + 2x} + \sqrt{3x} \big) } \:\: \: = \: k \:$

$\twoheadrightarrow \sf \: \: \dfrac{4\sqrt{a}}{3\times 2\sqrt{3a}} \:\: \: = \: k \:$

$\twoheadrightarrow \sf \: \: \dfrac{2\sqrt{a}}{3\times \sqrt{3a}} \:\: \: = \: k \:$

$\twoheadrightarrow \sf \: \: \dfrac{2}{3 \sqrt{3}} \:\: \: = \: k \:$

$\twoheadrightarrow \pmb {\underline {\boxed { \purple {\frak{ \: k\:=\:\: \dfrac{2}{3 \sqrt{3}} \:}}}}}\:\:\bigstar \:$

$\qquad \therefore \:\sf Hence \:k\:is \:Equals \:to \:\pmb{\sf Option \:A \:)\:\dfrac{2}{3\sqrt{3}}}\:. \:$

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\footnotesize\sf \: \lim\limits_{x\to a} \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} }$

If we substitute directly x = a, we get

$\rm \:  =  \: \dfrac{ \sqrt{a + 2a} - \sqrt{3a} }{ \sqrt{3a + a} - 2 \sqrt{a} }$

$\rm \:  =  \: \dfrac{ \sqrt{3a} - \sqrt{3a} }{ \sqrt{4a} - 2 \sqrt{a} }$

$\rm \:  =  \: \dfrac{ 0}{2 \sqrt{a} - 2 \sqrt{a} }$

$\rm \:  =  \: \dfrac{ 0}{0 } \: which \: is \: meaningless$

So, To evaluate this limit,

$\rm :\longmapsto\:\footnotesize\sf \: \lim\limits_{x\to a} \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} }$

On rationalizing the denominator and numerator, we get

$\rm \:  =  \: \footnotesize\sf \: \lim\limits_{x\to a} \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} } \times \dfrac{ \sqrt{3a + x} + 2 \sqrt{x} }{ \sqrt{3a + x} + 2 \sqrt{x} } \times \dfrac{ \sqrt{a + 2x} + \sqrt{3x} }{ \sqrt{a + 2x} + \sqrt{3x} }$

We know,

$\red{ \boxed{ \sf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

So, using this we get

$\rm \:  =  \: \displaystyle\lim_{x \to a}\sf \frac{[ \: (a + 2x) - 3x \: ] \: [ \: \sqrt{3a + x} + 2 \sqrt{x} \: ]}{[ \: 3a + x - 4x \: ] \: [ \: \sqrt{a + 2x} + \sqrt{3x} \: ] }$

$\rm \:  =  \: \displaystyle\lim_{x \to a}\sf \frac{[ \: a + 2x - 3x \: ] \: [ \: \sqrt{3a + x} + 2 \sqrt{x} \: ]}{[ \: 3a - 3x \: ] \: [ \: \sqrt{a + 2x} + \sqrt{3x} \: ] }$

$\rm \:  =  \: \displaystyle\lim_{x \to a}\sf \frac{[ \: a - x \: ] \: [ \: \sqrt{3a + x} + 2 \sqrt{x} \: ]}{[ \: 3a - 3x \: ] \: [ \: \sqrt{a + 2x} + \sqrt{3x} \: ] }$

$\rm \:  =  \: \displaystyle\lim_{x \to a}\sf \frac{[ \: a - x \: ] \: [ \: \sqrt{3a + x} + 2 \sqrt{x} \: ]}{[ \: 3(a - x) \: ] \: [ \: \sqrt{a + 2x} + \sqrt{3x} \: ] }$

$\rm \:  =  \: \displaystyle\lim_{x \to a}\sf \frac{ \: [ \: \sqrt{3a + x} + 2 \sqrt{x} \: ]}{3\: [ \: \sqrt{a + 2x} + \sqrt{3x} \: ] }$

$\rm \:  =  \: \sf \dfrac{ \: [ \: \sqrt{3a + a} + 2 \sqrt{a} \: ]}{3\: [ \: \sqrt{a + 2a} + \sqrt{3a} \: ] }$

$\rm \:  =  \: \sf \dfrac{ \: [ \: \sqrt{4a} + 2 \sqrt{a} \: ]}{3\: [ \: \sqrt{3a} + \sqrt{3a} \: ] }$

$\rm \:  =  \: \sf \dfrac{ \: [ \: 2\sqrt{a} + 2 \sqrt{a} \: ]}{3\: [ \: 2 \sqrt{3a} \: ] }$

$\rm \:  =  \: \sf \dfrac{ \: [ \: 4\sqrt{a} \: ]}{3\: [ \: 2 \sqrt{3a} \: ] }$

$\rm \:  =  \: \dfrac{2}{3 \sqrt{3} }$

So,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \:\footnotesize\sf \: \lim\limits_{x\to a} \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} } = \frac{2}{3 \sqrt{3} } }}}$

But it is given that,

$\rm :\longmapsto\:\green{ \boxed{ \sf{ \:\footnotesize\sf \: \lim\limits_{x\to a} \dfrac{ \sqrt{a + 2x} - \sqrt{3x} }{ \sqrt{3a + x} - 2 \sqrt{x} } = k }}}$

So, hence, we concluded that

$\red{ \boxed{ \sf{ \: \: \: \:k \: = \: \frac{2}{ \: \: 3 \sqrt{3} \: \: \: }}}}$

• Hence, Option (a) is correct.

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\sf \frac{sinx}{x} = 1}}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\sf \frac{tanx}{x} = 1}}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\sf \frac{log(1 + x)}{x} = 1}}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\sf \frac{ {e}^{x} - 1 }{x} = 1}}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0}\sf \frac{ {a}^{x} - 1 }{x} = loga}}}$