Question

limit problem

Answer 1

ok

Answer 2

lim x approaches - infinity of 2x divided by the square root of x squared + 1

Answer 3

\[\lim_{x \rightarrow \infty} \frac{2x}{\sqrt{x^2 + 1}}\]

Answer 4

negative infinity

Answer 5

oh

Answer 6

In any case, how would you make the terms in a form similar to 1/x (so you can get the limits)?

Answer 7

i'm not sure rationalize

Answer 8

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

Answer 9

Divide by x instead, but this question is a little ''tricky'', I think

Answer 10

Why divide by x?

Answer 11

\[\lim_{x \rightarrow -\infty} \frac{2x}{\sqrt{x^2 + 1}}\]\[=\lim_{x \rightarrow -\infty} \frac{\frac{2x}{x}}{\frac{\sqrt{x^2 + 1}}{x}}\]\[=-\lim_{x \rightarrow -\infty} \frac{2}{\sqrt{\frac{x^2 + 1}{x^2}}}\]\[=-\lim_{x \rightarrow -\infty} \frac{2}{\sqrt{1+\frac{1}{x^2}}}\]\[=-\frac{2}{\sqrt{1+0}}=...\]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 :|

Answer 12

oh, that's what i meant by sqrt(x^2) so it is x

Answer 13

I don't think you need a negative sign because it's x^2

Answer 14

No.. I think I need it (I've got similar question wrong for n times, so I remember it :\)

Answer 15

Oh, well the numerator is negative but the denominator is positive so it is negative