Limits at infinity

[mikagurl]

The problem is to evaluate: the limit as x approaches infinity.

.\(\displaystyle \displaystyle{\frac{\sqrt{3x^2+6}}{5-2x}}\)

The first thing I did was factored out and x because it was the largest power in the denominator. So now I have:

. . .\(\displaystyle \dfrac{x\, \sqrt{3x\, +\, \dfrac{6}{x}\,}}{\dfrac{5}{x}\, -\, 2}\)

The guide that I am using is saying to factor out an x^2 because we need an x in the numerator and in the denominator. This is where I am lost because we have an x in both the numerator and the denominator as shown in the last step. Maybe my last step is incorrect. Please help.

 

[stapel]

I have edited this thread to remove the lengthy conversation regarding what you'd intended your original posting to say. Now that we have most of that that initial information, let's continue:

The problem is to evaluate:

. . .\(\displaystyle \displaystyle{\frac{\sqrt{3x^2+6}}{5-2x}}\)

Your subject line refers to "Limits at infinity", so I will guess that the expression above was supposed to be a limit, taken as \(\displaystyle x\) gets arbitrarily large. In other words, the exercise is really this:

. . . . .\(\displaystyle \mbox{Evaluate }\, \displaystyle{\lim_{x\, \rightarrow \, \infty}}\, \displaystyle{\frac{\sqrt{3x^2\,+\,6\,}}{5\,-\,2x}}\)

The first thing I did was factored out and x because it was the largest power in the denominator. So now I have:

. . .\(\displaystyle \dfrac{x\, \sqrt{3x\, +\, \dfrac{6}{x}\,}}{\dfrac{5}{x}\, -\, 2}\)

It might help if you showed your steps. You seem to have factored \(\displaystyle \sqrt{x\,}\) out of the square root in the numerator, but you've only shown an \(\displaystyle x.\) Also, you appear to have factored an \(\displaystyle x\) out of the denominator, but it's disappeared. So I'm not at all clear on what you're doing.

Just as \(\displaystyle 2\, =\, \sqrt{2^2\,}\, =\, \sqrt{4\,},\) so also \(\displaystyle x\, =\, \sqrt{x^2\,}.\) So, in order to pull an \(\displaystyle x\) out front in the numerator, you just factor an \(\displaystyle x^2\) out of the contents of the square root. So try doing this first. Then factor (rather than "make disappear") an \(\displaystyle x\) from the expression in the denominator. This will give you something along the lines of:

. . . . .\(\displaystyle \displaystyle{ \lim_{x\, \rightarrow \, \infty} } \) \(\displaystyle \left(\displaystyle{ \frac{x\,\sqrt{\mbox{something}}}{x\left(\frac{5}{x}\, -\, 2\right)} }\right)\)

Then you can cancel, followed by taking the limit.

 

[mikagurl]

Oh. I think I get it. So you are saying that I need to factor out an x because it is the largest power in the denominator. But because we have a square root in the numerator, the only way to factor an x is by factoring an x^2??? Am I following you or missing what you are trying to say?:-?

 

[HallsofIvy]

I would do this a slightly different way. Since "infinity" is difficult to work with ("infinity" is not actually a number so you cannot substitute "infinity" for x) but 0 is, I would start by dividing both numerator and denominator by x:
\(\displaystyle \frac{\frac{1}{x}\sqrt{3x^2+ 6}}{\frac{1}{x}(5- 2x)}\)
Taking the \(\displaystyle \frac{1}{x}\) inside the square root makes it \(\displaystyle \frac{1}{x^2}\) this becomes
\(\displaystyle \frac{\sqrt{\frac{1}{x^2}(3x^2+ 6}}{\frac{1}{x}(5- 2x)}\)\(\displaystyle = \frac{\sqrt{3+ \frac{6}{x^2}}}{\frac{5}{x}- 2}\).

And, now, as x goes to infinity, those fractions with x in the denominator go to 0.

 

[markraz]

The problem is to evaluate: the limit as x approaches infinity.

.\(\displaystyle \displaystyle{\frac{\sqrt{3x^2+6}}{5-2x}}\)

The first thing I did was factored out and x because it was the largest power in the denominator. So now I have:

. . .\(\displaystyle \dfrac{x\, \sqrt{3x\, +\, \dfrac{6}{x}\,}}{\dfrac{5}{x}\, -\, 2}\)

The guide that I am using is saying to factor out an x^2 because we need an x in the numerator and in the denominator. This is where I am lost because we have an x in both the numerator and the denominator as shown in the last step. Maybe my last step is incorrect. Please help.

With these "limits at infinity" easiest thing to do is use "end behavior" from precal or use the following

Rules:

M<N => Limit = 0
M=N => Limit is the coefficients (which is what you have here)
M>N => Limit = +- INF (depending on sign of leading terms)

where:
M is highest exponent in Numerator
N is highest exponent in Denominator

So in your case you don't even have to calculate anything. You can see M=N therefore the limit is the coefficients -sqr(3)/2
If there are no coefficients then the limit would be 1, when M=N


IMO it's good to get in the habit of using this method since when you get to Calc II, with series, it comes in very very handy to easily
identify if a series goes to infinity. It can save a lot of time. also even if you use the longer method which may be preferred by your instructor...
Atleast you can easily check your answer quickly