finding horizontal asymptotes

[rioean17]

for the equation f(x)=2x/sqrt((2x^2)-1) i have found that the domain is (-infinity,infinity) and that there are no vertical asymptotes. also i understand the rules for finding horizontal asymptotes but because the denominator is a square root i don't understand how the rules would apply here. can some one EXPLAIN to me how to find the horizontal asymptotes. im not looking for someone to just give me the answer. i would like to know the thought process behind it.

 

[galactus]

\(\displaystyle \L\\\frac{2x}{\sqrt{2x^{2}-1}}\)


What about \(\displaystyle \L\\x=\frac{1}{\sqrt{2}}\)?.


What does f(x) approach as x approaches infinity or -infinity?.

 

[rioean17]

thank you, that helped alot. i think im starting to understand.

 

[stapel]

for the [function] f(x)=2x/sqrt((2x^2)-1)...can some one EXPLAIN to me how to find the horizontal asymptotes.

You've probably been given some sort of rule for horizontal asymptotes, where you compare the exponents on the leading terms. The square root, of course, messes that up here.

You've got:

. . . . .\(\displaystyle \L f(x)\, =\, \frac{2x}{\sqrt{2x^2\, -\, 1}}\)

We're going to fiddle with this by dividing through, top and bottom, by "x". Effectively, this is multiplying the fraction by (1/x)/(1/x), so we won't have changed anything. (Okay, we'll have messed with the domain a bit, but you're looking at the graph as x tends toward the far ends of the x-axis, so we can ignore the creation of a problem at x = 0.)

. . . . .\(\displaystyle \L \frac{\left(\frac{2x}{x}\right)} {\left(\frac{\sqrt{2x^2\, -\, 1}}{\sqrt{x^2}}\right)}\)

...because sqrt[x²] = x (for our purposes). Then:

. . . . .\(\displaystyle \L \frac{2}{\sqrt{\frac{2x^2\, - \,1}{x^2}}}\)

. . . . .\(\displaystyle \L \frac{2}{\sqrt{2\, -\, \frac{1}{x^2}}}\)

And, as x gets very large, you know that 1/x² gets very small, so you effectively have:

. . . . .\(\displaystyle \L \frac{2}{\sqrt{2\, - \, 0}\,=\, \frac{2}{\sqrt{2}}\, =\, \sqrt{2}\)

If you plug f(x) into your graphing calculator and ZOOM out or follow TABLE a ways out, you should start getting y-values that get closer and closer to 1.414...

i have found that the domain is (-infinity,infinity) and that there are no vertical asymptotes.

The x-value the tutor provided in the previous reply was in relation to the vertical asymptote: You'll have a vertical asymptote whenever the denominator is zero, so you need to solve 2x² - 1 = 0. This will confirm the value provided earlier.

Note also that you cannot have negatives inside a square root. This will bar part of the x-axis from the domain. Solve 2x² - 1 < 0 to find the part to omit.

Eliz.