Question

plz tell me the solution of the question.​

Answer 1

Step-by-step explanation:

We have,

$\lim_{x \rarr \infty } \dfrac{ \sqrt{ {x}^{2} + {a}^{2} } + \sqrt{ {x}^{2} + {b}^{2} } }{ \sqrt{ {x}^{2} + c^{2} } + \sqrt{ {x}^{2} + {d}^{2} } }$

$= \lim_{x \rarr \infty } \dfrac{ \sqrt{ {x}^{2} \bigg \{1+ { \bigg( \dfrac{a}{x} \bigg)}^{2} \bigg \} } + \sqrt{ {x}^{2} \bigg\{1+ { \bigg( \dfrac{b}{x} \bigg)}^{2} \bigg \} } }{ \sqrt{ {x}^{2} \bigg\{1+ { \bigg( \dfrac{c}{x} \bigg)}^{2} \bigg \} } + \sqrt{ {x}^{2} \bigg\{1+ { \bigg( \dfrac{d}{x} \bigg)}^{2} \bigg \} } }$

$= \lim_{x \rarr \infty } \dfrac{ x\sqrt{ 1+ { \bigg( \dfrac{a}{x} \bigg)}^{2} } + x\sqrt{ 1+ { \bigg( \dfrac{b}{x} \bigg)}^{2} } }{x \sqrt{ 1+ { \bigg( \dfrac{c}{x} \bigg)}^{2} } + x \sqrt{ 1+ { \bigg( \dfrac{d}{x} \bigg)}^{2} } }$

$= \lim_{x \rarr \infty } \dfrac{ \sqrt{ 1+ { \bigg( \dfrac{a}{x} \bigg)}^{2} } + \sqrt{ 1+ { \bigg( \dfrac{b}{x} \bigg)}^{2} } }{ \sqrt{ 1+ { \bigg( \dfrac{c}{x} \bigg)}^{2} } + \sqrt{ 1+ { \bigg( \dfrac{d}{x} \bigg)}^{2} } }$

$= \dfrac{ \sqrt{ 1+ { \bigg( \dfrac{a}{ \infty } \bigg)}^{2} } + \sqrt{ 1+ { \bigg( \dfrac{b}{ \infty } \bigg)}^{2} } }{ \sqrt{ 1+ { \bigg( \dfrac{c}{ \infty } \bigg)}^{2} } + \sqrt{ 1+ { \bigg( \dfrac{d}{ \infty } \bigg)}^{2} } }$

$= \dfrac{ \sqrt{ 1 } + \sqrt{ 1} }{ \sqrt{ 1 } + \sqrt{ 1} }$

$\sf{ \: = 1}$