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finaynay
 3 years ago
limit problem
finaynay
 3 years ago
limit problem

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finaynay
 3 years ago
Best ResponseYou've already chosen the best response.0lim x approaches  infinity of 2x divided by the square root of x squared + 1

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty} \frac{2x}{\sqrt{x^2 + 1}}\]

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0In any case, how would you make the terms in a form similar to 1/x (so you can get the limits)?

finaynay
 3 years ago
Best ResponseYou've already chosen the best response.0i'm not sure rationalize

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Divide by x instead, but this question is a little ''tricky'', I think

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0Why divide by x?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow \infty} \frac{2x}{\sqrt{x^2 + 1}}\]\[=\lim_{x \rightarrow \infty} \frac{\frac{2x}{x}}{\frac{\sqrt{x^2 + 1}}{x}}\]\[=\lim_{x \rightarrow \infty} \frac{2}{\sqrt{\frac{x^2 + 1}{x^2}}}\]\[=\lim_{x \rightarrow \infty} \frac{2}{\sqrt{1+\frac{1}{x^2}}}\]\[=\frac{2}{\sqrt{1+0}}=...\]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 :

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0oh, that's what i meant by sqrt(x^2) so it is x

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0I don't think you need a negative sign because it's x^2

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1No.. I think I need it (I've got similar question wrong for n times, so I remember it :\)

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, well the numerator is negative but the denominator is positive so it is negative
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