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The limit: $ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $

(A) is 0 (B) is $\frac 1 2 $ (C) is 2 (D) does not exist

Is doing it with Binomial expansion and cancelling the terms only way?

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3 Answers 3

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$$ \lim _{x \to \infty } (\sqrt{x^2 +x} - \sqrt{x^2 +1} )=\lim _{x \to \infty } (\sqrt{x^2 +x} - \sqrt{x^2 +1} )\frac{\sqrt{x^2 +x} + \sqrt{x^2 +1}}{\sqrt{x^2 +x} + \sqrt{x^2 +1}}=\lim _{x \to \infty } \frac{x-1}{\sqrt{x^2 +x} + \sqrt{x^2 +1}}=...=1/2$$

(if you need more I could finish it)

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  • $\begingroup$ Wait let me do!! :( $\endgroup$
    – N S
    Mar 17, 2015 at 14:49
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    $\begingroup$ But done already.. Finish it... $\endgroup$
    – N S
    Mar 17, 2015 at 14:49
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$$ \lim \limits_{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1}=\lim \limits_{x \to \infty }\dfrac{(\sqrt{x^2 +x} - \sqrt{x^2 +1})(\sqrt{x^2 +x} + \sqrt{x^2 +1})}{\sqrt{x^2+x}+\sqrt{x^2+1}}=\lim \limits_{x \to \infty } \dfrac{x-1}{\sqrt{x^2+x}+\sqrt{x^2+1}}=\lim \limits_{x \to \infty } \dfrac{1-\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x}}+\sqrt{1+\dfrac{1}{x^2}}}=\dfrac{1}{2} $$

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\begin{array} \\ \displaystyle \lim_{x \rightarrow \infty} \sqrt{x^2 + x} - \sqrt{x^2 + 1} &= \displaystyle \lim_{x \rightarrow \infty} \left(\sqrt{x^2 + x} - \sqrt{x^2 + 1}\right) \frac{\sqrt{x^2 + x} + \sqrt{x^2 + 1}}{\sqrt{x^2 + x} + \sqrt{x^2 + 1}} \\ &= \displaystyle \lim_{x \rightarrow \infty} \frac{(x^2 + x) - (x^2 + 1)}{\sqrt{x^2 + x} + \sqrt{x^2 + 1}} \\ &= \displaystyle \lim_{x \rightarrow \infty} \frac{x-1}{\sqrt{x^2 + x} + \sqrt{x^2 + 1}} \cdot \frac{(1/x)}{(1/x)} \\ &= \displaystyle \lim_{x \rightarrow \infty} \frac{1-1/x}{\sqrt{1 + 1/x} + \sqrt{1 + 1/x^2}} \\ &= \displaystyle \frac{1}{2} \end{array}

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