Question

Lim x tends to zero a minus roor a square minus x square by x square

Answer 1

Given Expression,

$\displaystyle \sf \lim_{ x \to \: 0} \dfrac{a - \sqrt{ {a}^{2} - { {x}^{2} }^{} } }{ {x}^{2} }$

Dividing and Multiplying by a + √a² + x².

$\longrightarrow \sf \: \displaystyle \sf \lim_{ x \to \: 0} \: \: \dfrac{a - \sqrt{ {a}^{2} - { {x}^{2} }^{} } }{ {x}^{2} } \times \dfrac{a + \sqrt{a {}^{2} - {x}^{2} } }{a + \sqrt{ {a}^{2} - {x}^{2} } } \\ \\ \longrightarrow \sf \: \displaystyle \sf \lim_{ x \to \: 0} \: \: \dfrac{a {}^{2} - (\sqrt{ {a}^{2} - { {x}^{2} }^{} }) {}^{2} }{ {x}^{2} (a + \sqrt{ {a}^{2} - {x}^{2} } )} \\ \\ \longrightarrow \sf \: \displaystyle \sf \lim_{ x \to \: 0} \: \: \dfrac{a {}^{2} - ( {a}^{2} - x {}^{2} ) }{ {x}^{2} (a + \sqrt{ {a}^{2} - {x}^{2} } )} \\ \\ \longrightarrow \sf \: \: \displaystyle \sf \lim_{ x \to \: 0} \: \: \dfrac{a {}^{2} - {a}^{2} + x {}^{2} }{ {x}^{2} (a + \sqrt{ {a}^{2} - {x}^{2} })} \\ \\ \longrightarrow \sf \: \: \displaystyle \sf \lim_{ x \to \: 0} \: \: \dfrac{x {}^{2} }{ {x}^{2} (a + \sqrt{ {a}^{2} - {x}^{2} })} \\ \\ \longrightarrow \sf \: \: \displaystyle \sf \lim_{ x \to \: 0} \: \dfrac{1}{a + \sqrt{ {a}^{2} - {x}^{2} }}$

Putting x = 0,

$\longrightarrow \sf \: \dfrac{1}{ {a}^{} + \sqrt{ {a}^{2} + 0} } \\ \\ \longrightarrow \sf \: \dfrac{1}{ {a}^{} + a } \\ \\ \longrightarrow \sf \: \dfrac{1}{2a }$

Thus,

$\displaystyle \sf \lim_{ x \to \: 0} \: \dfrac{a - \sqrt{ {a}^{2} - { {x}^{2} }^{} } }{ {x}^{2} } = \dfrac{1}{2a}$