Math, asked by sumit513, 11 months ago

solve this ???I want to know the answer​

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Answers

Answered by pulakmath007
1

SOLUTION

TO EVALUATE

\displaystyle  \sf{\lim_{x \to 0 \: }  \:  \frac{ \sqrt{1 +  {x}^{2}  } -  \sqrt{1 + x}  }{ \sqrt{1 +  {x}^{3} } -  \sqrt{1 + x}  } }

EVALUATION

\displaystyle  \sf{\lim_{x \to 0 \: }  \:  \frac{ \sqrt{1 +  {x}^{2}  } -  \sqrt{1 + x}  }{ \sqrt{1 +  {x}^{3} } -  \sqrt{1 + x}  } }

\displaystyle  \sf{ = \lim_{x \to 0 \: }  \:  \frac{( \sqrt{1 +  {x}^{2}  } -  \sqrt{1 + x} )( \sqrt{1 +  {x}^{2}  }  +  \sqrt{1 + x} ) (\sqrt{1 +  {x}^{3} }  +   \sqrt{1 + x}  )}{ (\sqrt{1 +  {x}^{3} }  +  \sqrt{1 + x}  )(\sqrt{1 +  {x}^{3} } -  \sqrt{1 + x}  )( \sqrt{1 +  {x}^{2}  }  +   \sqrt{1 + x} )} }

\displaystyle  \sf{ = \lim_{x \to 0 \: }  \:  \frac{( 1 +  {x}^{2}   -  1  - x ) (\sqrt{1 +  {x}^{3} }  +   \sqrt{1 + x}  )}{ (1 +  {x}^{3}  -   1  - x )( \sqrt{1 +  {x}^{2}  }  +   \sqrt{1 + x} )} }

\displaystyle  \sf{ = \lim_{x \to 0 \: }  \:  \frac{(  {x}^{2}  - x ) (\sqrt{1 +  {x}^{3} }  +   \sqrt{1 + x}  )}{ ( {x}^{3}  - x )( \sqrt{1 +  {x}^{2}  }  +   \sqrt{1 + x} )} }

\displaystyle  \sf{ = \lim_{x \to 0 \: }  \:  \frac{x( x - 1 ) (\sqrt{1 +  {x}^{3} }  +   \sqrt{1 + x}  )}{ x( {x}^{2}  - 1 )( \sqrt{1 +  {x}^{2}  }  +   \sqrt{1 + x} )} }

\displaystyle  \sf{ = \lim_{x \to 0 \: }  \:  \frac{x( x - 1 ) (\sqrt{1 +  {x}^{3} }  +   \sqrt{1 + x}  )}{ x( x + 1)(x  -  1)( \sqrt{1 +  {x}^{2}  }  +   \sqrt{1 + x} )} }

\displaystyle  \sf{ = \lim_{x \to 0 \: }  \:  \frac{x(\sqrt{1 +  {x}^{3} }  +   \sqrt{1 + x}  )}{ (x + 1)( \sqrt{1 +  {x}^{2}  }  +   \sqrt{1 + x} )} }

\displaystyle  \sf{ =  \:  \frac{0(\sqrt{1 +  {0}^{3} }  +   \sqrt{1 + 0}  )}{ (0 + 1)( \sqrt{1 +  {0}^{2}  }  +   \sqrt{1 + 0} )} }

\displaystyle  \sf{ =   \frac{0 \times 2}{1 \times 2} }

\displaystyle  \sf{ =  0 }

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