Question

limit X approaches to zero x divided by 3-
$\sqrt{x + 9}$
​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \: \dfrac{x}{3 - \sqrt{x + 9} }$

On directly Substitute x = 0, we get

$\rm \: = \: \: \dfrac{0}{3 - \sqrt{0 + 9} }$

$\rm \: = \: \: \dfrac{0}{3 - \sqrt{ 9} }$

$\rm \: = \: \: \dfrac{0}{3 - 3}$

$\rm \: = \: \: \dfrac{0}{0}$

$\rm \: = \: \: which \: is \: meaningless$

So, Consider again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \: \dfrac{x}{3 - \sqrt{x + 9} }$

$\rm \: = \: \: \:\displaystyle\lim_{x \to 0}\rm \: \dfrac{x}{3 - \sqrt{x + 9} } \times \dfrac{3 + \sqrt{x + 9} }{3 + \sqrt{x + 9} }$

$\rm \: = \: \: \:\displaystyle\lim_{x \to 0}\rm \: \dfrac{x(3 + \sqrt{x + 9} \: ) }{ {3}^{2} - (x + 9)}$

$\rm \: = \: \: \:\displaystyle\lim_{x \to 0}\rm \: \dfrac{x(3 + \sqrt{x + 9} \: ) }{9 - (x + 9)}$

$\rm \: = \: \: \:\displaystyle\lim_{x \to 0}\rm \: \dfrac{x(3 + \sqrt{x + 9} \: ) }{9 - x - 9}$

$\rm \: = \: \: \:\displaystyle\lim_{x \to 0}\rm \: \dfrac{x(3 + \sqrt{x + 9} \: ) }{ - x}$

$\rm \: = \: \: \:\displaystyle\lim_{x \: \to \: 0}\rm \: \dfrac{(3 + \sqrt{x + 9} \: ) }{ - 1}$

$\rm \: = \: \: - (3 + \sqrt{9 + 0} \: )$

$\rm \: = \: \: - (3 + \sqrt{9} \: )$

$\rm \: = \: \: - (3 + 3 \: )$

$\rm \: = \: \: - 6$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\bf \: \dfrac{x}{3 - \sqrt{x + 9} } = - \: 6$

Additional Information :-

$\boxed{ \bf{ \displaystyle\lim_{x \to 0}\bf\: \dfrac{sinx}{x} = 1}}$

$\boxed{ \bf{ \displaystyle\lim_{x \to 0}\bf\: \dfrac{tanx}{x} = 1}}$

$\boxed{ \bf{ \displaystyle\lim_{x \to 0}\bf\: \dfrac{ {sin}^{ - 1} x}{x} = 1}}$

$\boxed{ \bf{ \displaystyle\lim_{x \to 0}\bf\: \dfrac{ {tan}^{ - 1} x}{x} = 1}}$

$\boxed{ \bf{ \displaystyle\lim_{x \to 0}\bf\: \dfrac{ log(1 + x) }{x} = 1 }}$

$\boxed{ \bf{ \displaystyle\lim_{x \to 0}\bf\: \dfrac{ {e}^{x} - 1}{x} = 1 }}$

$\boxed{ \bf{ \displaystyle\lim_{x \to 0}\bf\: \dfrac{ {a}^{x} - 1}{x} = loga }}$