Math, asked by satwikdatta728, 6 months ago

please answer the question please​

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Answered by HermioneJeanGranger1
1

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solution to the question given above

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Answered by SrijanShrivastava
1

 \\  \lim  _{x →0}  [\frac{ \sqrt{7 + x} -  \sqrt{7  + {x}^{2} }  }{ \sqrt{7 -  {x}^{2}  }  -  \sqrt{ 7 - x} } ]≡ \frac{0}{0}

Using L'Hôspital's rule of Limits

 \\  \lim_{x→0}  \frac{ \frac{1}{2 \sqrt{7 + x} } -  \frac{2x}{ 2\sqrt{7 +  {x}^{2} } }  }{  - \frac{2x}{2 \sqrt{7 -  {x}^{2} } }   + \frac{1}{ 2\sqrt{7 - x}  } }

 \\  \lim  _{x →0} \frac{( \sqrt{7 +  {x}^{2}  }  - 2x \sqrt{7 + x} ) \sqrt{(7 -  {x}^{2} )(7 - x)} }{( - 2x \sqrt{7 - x} +  \sqrt{7 -  {x}^{2} }  ) \sqrt{(7 + x)(7 +  {x}^{2} ) } }

 \\  \lim  _{x →0} \frac{ \sqrt{343 - 49x - 7 {x}^{4} +  {x}^{5}   } - 2x \sqrt{343 - 56 {x}^{2} +  {x}^{4}  } }{( \sqrt{7 -  {x}^{2}  - 2x \sqrt{7 - x} } ) \sqrt{49 + 7x + 7 {x}^{2} +  {x}^{3}  } }

 \\  \implies   \lim  _{x →0}  \frac{ \sqrt{7 + x} -  \sqrt{7  + {x}^{2} }  }{ \sqrt{7 -  {x}^{2}  }  -  \sqrt{ 7 - x} }  → (1 )=  \frac{ \sqrt{343} }{  \sqrt{7} \sqrt{49} }

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