Evaluating a limit at infinity

[peblez]

This is a textbook example, but i am not unsure on one of the steps they did and am in need of clarification on why they did what they did

Find the limit of this:


lim x --> infinity [ sqrt ( 2x^2 + 1) ] / 3x - 5

It says divide both numerator and denominator by x

= lim x --> infinity [ sqrt ( 2 + 1/ x^2 ) ] / ( 3 - 5/ x) ( since sqrt (x^2) = x for > 0)


Can someone explain those two steps to me. I don't understand how sqrt(2x^2 + 1 ) became sqrt ( 2 + 1/x^2) when they are only dividing by x.



Also another example i am confused about is this, its the same equation just that the limit is now x --> - infinity

lim x --> - infinity [sqrt ( 2x^2 + 1) / ( 3x - 5)

They say " In computing the limit as x --> - infinity, we must remember that for x < 0, we have sqrt(x^2) = | x | = - x. So when we divide the numerator by x, for x < 0 we get.

1/x ( sqrt ( 2x^2+1) = - 1/ (sqrt( x^2) ( sqrt ( 2x^2 + 1 ) = - ( sqrt ( 2 + 1 / x^2)

Can some one explain those 3 processes to me, i don't understand why they have to make x = sqrt ( x^2), and i dont' understand exactly where the negative sign came from.


thanks

 

[Loren]

\(\displaystyle \frac{\sqrt{2x^2 +1}}{3x-5}=\frac{\frac{1}{x}\sqrt{2x^2+1}}{\frac{1}{x}(3x-5)}=\frac{\sqrt{\frac{2x^2+1}{x^2}}}{\frac{3x-5}{x}}=\frac{\sqrt{2+\frac{1}{x^2}}}{3-\frac{5}{x}}\)

 

[peblez]

Ah i see.

But can you explain the one on negative infinity

 

[o_O]

\(\displaystyle \lim_{x \to -\infty}\frac{\sqrt{2x^{2}+1}}{3x - 5}\)

Thing to note with problems like these involving negative infinity and square roots. First, note that \(\displaystyle \sqrt{x^{2}} = |x|\). Since x is approaching NEGATIVE infinity, we can presume x to be negative. This means that IN THIS PROBLEM, x can be expressed as a square root: \(\displaystyle x = -\sqrt{x^2}\) where the negative sign is there because \(\displaystyle \sqrt{x^{2}} = |x|\) is positive. Hmm, don't know if this is explained clear enough.

 

[pka]

The ole analyst must step in. Clearly if the limit exists, it must be negative.
However, the solution that Loren gives is a positive limit.
Here is the falsity there. If \(\displaystyle x < 0\quad \Rightarrow \quad \frac{x}{{\left| x \right|}} = - 1\).
As 0_o notes \(\displaystyle \sqrt {x^2 } = \left| x \right|\).

 

[galactus]

\(\displaystyle \L\\\lim_{x\to-\{\infty}}\frac{\sqrt{2x^{2}+1}}{3x-5}\)

Because \(\displaystyle |x|=\sqrt{x^{2}}\), then you can use -x where convenient. Since the limit is clearly negative:

If you divide the bottom by negative x, then:

\(\displaystyle \L\\\lim_{x\to\{-\infty}}\frac{\sqrt{2+\frac{1}{x^{2}}}}{-3-\frac{5}{x}}=\frac{-\sqrt{2}}{3}\)


EDIT: pka already confirmed this. Sorry.