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lim x approaches - infinity of 2x divided by the square root of x squared + 1
\[\lim_{x \rightarrow \infty} \frac{2x}{\sqrt{x^2 + 1}}\]

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Other answers:

negative infinity
oh
In any case, how would you make the terms in a form similar to 1/x (so you can get the limits)?
i'm not sure rationalize
I wouldn't bother rationalizing the denominator. Waste of time. Try dividing by the least common multiple...I think that's what it's called? idk but it's x^2 so divide both numerator and denominator by x^2
Divide by x instead, but this question is a little ''tricky'', I think
Why divide by x?
\[\lim_{x \rightarrow -\infty} \frac{2x}{\sqrt{x^2 + 1}}\]\[=\lim_{x \rightarrow -\infty} \frac{\frac{2x}{x}}{\frac{\sqrt{x^2 + 1}}{x}}\]\[=-\lim_{x \rightarrow -\infty} \frac{2}{\sqrt{\frac{x^2 + 1}{x^2}}}\]\[=-\lim_{x \rightarrow -\infty} \frac{2}{\sqrt{1+\frac{1}{x^2}}}\]\[=-\frac{2}{\sqrt{1+0}}=...\]I'm not quite sure since when I should put that -ve sign. But if I remember correctly, I have to put a -ve sign :|
oh, that's what i meant by sqrt(x^2) so it is x
I don't think you need a negative sign because it's x^2
No.. I think I need it (I've got similar question wrong for n times, so I remember it :\)
Oh, well the numerator is negative but the denominator is positive so it is negative

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