Question

evaluate this limit. ​

Answer 1

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

Answer 2

$\huge\mathfrak\red{answer}$

look at the pic.......

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