Question

$\rm \lim_{x \to \sqrt{10} } \: \frac{ \sqrt{7 - 2x} - ( \sqrt{5} - \sqrt{2} )}{ {x}^{2} - 10}$


Evaluate !!​

Answer 1

ANSWER:

$\displaystyle \sf \lim_{x \to \sqrt{10}} \dfrac{ \sqrt{7-2x} - ( \sqrt{5}- \sqrt{2})}{{x}^{2} -10}= \dfrac{ - (\sqrt{5} + \sqrt{2})}{6\sqrt{10}}$

TO FIND:

$\displaystyle \sf \lim_{x \to \sqrt{10}} \dfrac{ \sqrt{7-2x} - ( \sqrt{5}- \sqrt{2})}{{x}^{2} -10}$

EXPLANATION:

$\implies \displaystyle \sf \lim_{x \to \sqrt{10}} \dfrac{ \sqrt{7-2x} - ( \sqrt{5}- \sqrt{2})}{{x}^{2} -10} \\ \\ \\ \implies\sf \dfrac{ \sqrt{7-2 \sqrt{10} } - ( \sqrt{5}-\sqrt{2})}{{ (\sqrt{10}) }^{2}-10} \\ \\ \\ \implies\sf \dfrac{ \sqrt{ 5 + 2-2 \sqrt{5} \sqrt{2} } - ( \sqrt{5}- \sqrt{2})}{10 -10} \\ \\ \\ \implies\sf \dfrac{ \sqrt{ (\sqrt{5} - \sqrt{2} {)}^{2} } - ( \sqrt{5} - \sqrt{2} ) }{10-10}$

$\implies\sf \dfrac{(\sqrt{5} - \sqrt{2} )- ( \sqrt{5} - \sqrt{2} ) }{10 -10} \\ \\ \\ \implies\sf \dfrac{0}{0} = \infty \\ \\ \\ \implies \sf Apply \ L'\ Hospital's \ rule \\ \\ \\ \implies \displaystyle \sf \lim_{x \to \sqrt{10}} \dfrac{ \dfrac{d}{dx} \left(\sqrt{7-2x} - ( \sqrt{5}- \sqrt{2}) \right)}{ \dfrac{d}{dx}( {x}^{2}-10)} \\ \\ \\ \implies\bigstar\ \boxed{ \bold{ \large{ \blue{ \dfrac{d}{dx} {x}^{n} = n {x}^{n - 1} }}}}$

$\implies \displaystyle \sf \lim_{x \to \sqrt{10}} \dfrac{ \dfrac{1}{ 2\sqrt{7 - 2x} } \dfrac{d}{dx} \left(7-2x \right) - (0 - 0)}{ 2x - 0} \\ \\ \\ \implies \displaystyle \sf \lim_{x \to \sqrt{10}} \dfrac{ \dfrac{ - 2}{ 2\sqrt{7 - 2x} } }{ 2x} \\ \\ \\ \implies \displaystyle \sf \lim_{x \to \sqrt{10}} \ \dfrac{ - 2}{4x \sqrt{7 - 2x} }$

$\implies \displaystyle \sf \lim_{x \to \sqrt{10}} \ \dfrac{ - 1}{2x \sqrt{7 - 2x} } \\ \\ \\ \implies\sf \dfrac{ - 1}{2 \sqrt{10} \sqrt{7 - 2 \sqrt{10} } } \\ \\ \\ \implies\sf \dfrac{ - 1}{2 \sqrt{10} \sqrt{5 + 2-2 \sqrt{5} \sqrt{2}} } \\ \\ \\ \implies\sf \dfrac{ - 1}{2 \sqrt{10} \sqrt{( \sqrt{5} - \sqrt{2} {)}^{2} }}$

$\implies\sf \dfrac{ - 1}{2 \sqrt{10}( \sqrt{5} - \sqrt{2} )}\\ \\ \\ \implies\sf \dfrac{ - 1}{2 \sqrt{10}( \sqrt{5} - \sqrt{2} )} \times \dfrac{\sqrt{5} + \sqrt{2} }{\sqrt{5} + \sqrt{2} } \\ \\ \\ \implies\sf \dfrac{ - (\sqrt{5} + \sqrt{2})}{2 \sqrt{10}( (\sqrt{5} {)}^{2} - (\sqrt{2} {)}^{2} )}$

$\implies\sf \dfrac{ - (\sqrt{5} + \sqrt{2})}{2 \sqrt{10}(5 - 2)} \\ \\ \\ \implies\sf \dfrac{ - (\sqrt{5} + \sqrt{2})}{6\sqrt{10}} \\ \\ \\ \implies \displaystyle \sf \lim_{x \to \sqrt{10}} \dfrac{ \sqrt{7-2x} - ( \sqrt{5}- \sqrt{2})}{{x}^{2} -10}=\dfrac{ - (\sqrt{5} + \sqrt{2})}{6\sqrt{10}}$

L' Hospital's rule:

• When ever the value of intermediate comes as 0/0 or ∞/∞, we use L' Hospital's rule.

• L' Hospital's rule is simply differentiating both numerator and denominator separately.

NOTE : Refer for another method in attachment.