Question

lim x tends to 0 (sqrt(a^2+x^2))-(sqrt(a^2-x^2))/x^2​

Answer 1

Step-by-step explanation:

We have,

$\lim _{x \rarr0 } \frac{ \sqrt{ {a}^{2} + {x}^{2} } - \sqrt{ {a}^{2} - {x}^{2} } }{ {x}^{2} }$

$= \lim_{x \rarr0} \frac{( \sqrt{ {a}^{2} + {x}^{2} } - \sqrt{ {a}^{2} - {x}^{2} } )( \sqrt{ {a}^{2} + {x}^{2} } + \sqrt{ {a}^{2} - {x}^{2} }) }{ {x ^{2} ( \sqrt{ {a}^{2} + {x}^{2} } + \sqrt{ {a}^{2} - {x}^{2} }) } }$

$= \lim_{x \rarr0} \frac{ {a}^{2} + {x}^{2} - {a}^{2} + {x}^{2} }{ {x}^{2} ( \sqrt{ {a}^{2} + {x}^{2} } + \sqrt{ {a}^{2} - {x}^{2} }) }$

$= \lim_{x \rarr0} \frac{2 {x}^{2} }{ {x}^{2} ( \sqrt{ {a}^{2} + {x}^{2} } + \sqrt{ {a}^{2} - {x}^{2} })}$

$= \lim_{x \rarr0} \frac{2}{ \sqrt{ {a}^{2} + {x}^{2} } + \sqrt{ {a}^{2} - {x}^{2} } }$

$= \frac{2}{a + a}$

$= \frac{1}{a}$