Question

limit x tends to 3 x-3 by square root of x-2 - square root of 4-x

Answer 1

EXPLANATION.

$\sf \implies \: lim \: _{x \to \: 3} \: = \dfrac{x \: - \: 3}{ \sqrt{x - 2} - \sqrt{4 - x} }$

$\sf \implies \: put \: the \: value \: of \: x \: = 3 \: in \: equation \\ \\ \sf \implies \: \frac{3 - 3}{ \sqrt{3 - 2} - \sqrt{4 - 3} } \\ \\\sf \implies \: \frac{0}{ \sqrt{1} - \sqrt{1} } \\ \\ \sf \implies \: it \: is \: in \: the \: form \: of \: \frac{0}{0}$

$\sf \implies \: if \: root \: \: is \: in \: equation \: we \: can \: \: simply \: \: rationalise \\ \\ \sf \implies \: lim_{x \to \: 3} \: \frac{x - 3}{ \sqrt{x - 2} - \sqrt{4 - x} } \: \times \: \frac{ \sqrt{x - 2} + \sqrt{4 - x} }{ \sqrt{x - 2} + \sqrt{4 - x} } \\ \\ \sf \implies \: lim_{x \to \: 3} \: = \frac{(x - 3)( \sqrt{x - 2} + \sqrt{4 - x} )}{( \sqrt{x - 2}) {}^{2} - ( \sqrt{4 - x} ) {}^{2} } \\ \\ \sf \implies \: \lim_{x \to \: 3} \: = \frac{(x - 3)( \sqrt{x - 2} + \sqrt{4 - x}) }{(x - 2 - (4 - x))}$

$\sf \implies \: \lim_{x \: \to \: 3} \: = \dfrac{(x + 3)( \sqrt{x - 2} + \sqrt{4 - x} )}{x - 2 - 4 + x} \\ \\ \sf \implies \: \lim_{x \: \to \: 3} \: = \frac{(x - 3)( \sqrt{x - 2} + \sqrt{4 - x} )}{2x - 6} \\ \\ \sf \implies \: \lim_{x \: \to \: 3} \: = \frac{(x - 3)( \sqrt{x - 2} + \sqrt{4 - x} )}{2(x - 3)}$

$\sf \implies \: \lim_{x \: \to \: 3} \: = \dfrac{ \sqrt{x - 2} + \sqrt{4 - x} }{2} \\ \\ \sf \implies \: put \: the \: value \: of \: x = 3 \: \\ \\ \sf \implies \: \lim_{x \: \to \: 3} \: = \frac{ \sqrt{3 - 2} + \sqrt{4 - 3} }{2} \\ \\ \sf \implies \: \lim_{x \: \to \: 3} \: = \frac{1}{2} = answer$