Math, asked by BrainlyKingBoss, 2 months ago

evaluate this limit. ​

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Answered by Anonymous
7

Explanation,

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\green\bigstar \:  \tt \:  lim_{x \rightarrow \:   \infty } \bigg(\sqrt{x {}^{2}  + x + 1}  -  \sqrt{x {}^{2}   - x - 1}  \bigg)

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 \leadsto \tt \:  \tt \:  lim_{x \rightarrow \:   \infty }  \bigg(\sqrt{x {}^{2}  + x + 1}  -  \sqrt{x {}^{2}   - x - 1}   \bigg) \times  \bigg(  \dfrac{\sqrt{x {}^{2}    +    x    +    1 } \:  +  \: \sqrt{x {}^{2}   -   x   -   1 }   }{\sqrt{x {}^{2}    +    x    +   1 }  + \sqrt{x {}^{2}   -   x   -   1 } } \bigg)

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\tt \leadsto \: lim_{x \rightarrow \:   \infty }  \bigg(  \dfrac{( \sqrt{x {}^{2} + x + 1 }  -  \sqrt{x {}^{2}  + x + 1}   ) \times (  \sqrt{x {}^{2} + x + 1 }  + \sqrt{x {}^{2}  - x - 1}  ) }{ \sqrt{x {}^{2}  + x + 1} +\sqrt{x {}^{2}  - x - 1}   } \bigg)

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\leadsto \tt \:  lim_{x \rightarrow \:  \infty }\bigg( \dfrac{ \not{x {}^{2}}+ x + 1 -  \cancel{x {}^{2}} + x + 1  }{ \sqrt{ x {}^{2} + x + 1 }  -  \sqrt{x {}^{2}   -  x  -  1} }  \bigg)

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 \leadsto \tt \:  lim_{x \rightarrow \:  \infty } \bigg( \dfrac{2x + 2}{ \sqrt{x {}^{2} + x + 1 }  -  \sqrt{x {}^{2} - x - 1 } }   \bigg)

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 \tt  \leadsto \:  lim _{x \rightarrow \:  \infty }   \dfrac{\not{x} \times  \bigg (2 +   \dfrac{2}{ x} \bigg)}{    \bigg(  \sqrt{ 1 +  \dfrac{ 1 }{x} +  \dfrac{1}{x {}^{2} }  }    \bigg)  +   \bigg(\sqrt{1 -  \dfrac{1}{x} -  \dfrac{1}{x {}^{2} }  } \bigg) }

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\tt  \leadsto \:  lim _{x \rightarrow \:  \infty }   \dfrac{\bigg (2 +   \dfrac{2}{ x} \bigg)}{     \bigg(  \sqrt{ 1 +  \dfrac{ 1 }{x} +  \dfrac{1}{x {}^{2} }  }    \bigg)  +   \bigg(\sqrt{1 -  \dfrac{1}{x} -  \dfrac{1}{x {}^{2} }  } \bigg) }

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\leadsto \tt \:   \dfrac{2 + 2 \times 0}{ \sqrt{1 + 0 + 0}   +   \sqrt{1  - 0 - 0} }

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\leadsto \tt   \not\dfrac{2}{2}

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\leadsto \:    {\underline{\tt{ \boxed{1}}} }\:  \red\bigstar

Answered by Anonymous
2

\huge\mathfrak\red{answer}

look at the pic.......

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