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$$\lim_{x\rightarrow\pm\infty} \frac{2x}{2x-\sqrt{4x^2-2x}}$$

What I did

I multiplied by the denominator's conjugate and got the following $$2x+\sqrt{4x^2-2x}$$

My question is, what would I now do to evaluate the limit? The positive infinity I would do as follows:

$$\lim_{x\rightarrow\infty}2x+\sqrt{4x^2-2x} = 2\cdot\infty + \sqrt{4\cdot(\infty)^2 - 2\cdot\infty} = \infty???$$

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  • $\begingroup$ but why this error occures there other questions in limits where i find errors while calculating $\endgroup$ Aug 17, 2021 at 2:00

1 Answer 1

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When $x \to +\infty$

$$2x+\sqrt{4x^2-2x} = 2x\left(1 +\sqrt{1- \dfrac{1}{2x}}\right) \sim 4x\to +\infty$$

When $x \to -\infty$

$$2x+\sqrt{4x^2-2x} = 2x\left(1 -\sqrt{1- \dfrac{1}{2x}}\right) = \dfrac{1}{1 +\sqrt{1- \dfrac{1}{2x}}} \to \dfrac{1}{2}$$

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  • $\begingroup$ Why does the sign change inside the paranthesis when you factored the $2x$? Nvm, I see why. Thank you ! $\endgroup$
    – B. Lee
    Nov 3, 2014 at 19:36

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