Question

Class 11th Limits CBSE ​

Answer 1

Step-by-step explanation:

$\lim_{x \rarr0} \frac{ \sqrt{ {x}^{2} + 16 } - 4}{ \sqrt{ {x}^{2} + 9} - 3}$

$= \lim_{x \rarr0} \frac{( \sqrt{ {x}^{2} + 16 } - 4)( \sqrt{ {x}^{2} + 16} + 4) ( \sqrt{ {x}^{2} + 9 } + 3)}{( \sqrt{ {x}^{2} + 9} - 3)( \sqrt{ {x}^{2} + 9} + 3)( \sqrt{ {x}^{2} + 16 } + 4) }$

$\lim_{x \rarr0} \frac{ ({x}^{2} + 16 - 16)( \sqrt{ {x}^{2} + 9} + 3) }{( {x}^{2} + 9 - 9)( \sqrt{ {x}^{2} + 16} + 4)}$

$\lim_{x \rarr0} \frac{ {x}^{2}( \sqrt{ {x}^{2} + 9} + 3)}{ {x}^{2} ( \sqrt{ {x}^{2} + 16} + 4)}$

$\lim_{x \rarr0} \frac{ \sqrt{ {x}^{2} + 9} + 3}{ \sqrt{ {x}^{2} + 16} + 4}$

$= \frac{6}{8} = \frac{3}{4}$