Limit as X approaches infinity and Asymptotes

[jpnov]

I'm not sure at all how to solve this one. Square root is giving me trouble. Can someone please give me some tips on how to start it?

This one is also giving me trouble because of the square root in the denominator. A vertical asymptote comes from when the denominator equals 0, but I tried setting the denom. = to zero and solving, but I couldn't factor. Also, how do you find the horizontal asymptotes - normally I would divide everything by the highest power of x, but I don't know how to do that with the sqrt in the bottom.

 

[Ryan Rigdon]

i hope this helps with 24.

 

[jpnov]

Thank you. How did you get 2 in the denominator?

 

[jpnov]

Nvm, got 24. Thanks

 

[jpnov]

Still need help with 34... if anyone has the time please. Thank you!

 

[mmm4444bot]

This one is also giving me trouble because of the square root in the denominator.

I'll show you an approach (below).

BTW: The instructions are somewhat misleading. They should read, "Find any horizontal or vertical asymptotes".

I tried setting the denom. = to zero and solving, but I couldn't factor.

When you can't solve a quadratic equation by factoring, use the Quadratic Formula.

Hint: If you're allowed to use a graphing calculator -- as a tool -- in these exercises, then plot the graph of the radicand, and check whether or not it intersects the x-axis. Doing that would also confirm whether or not function F has any vertical asymptotes.

Here's an approach, for finding the horizontal asymptotes. Square the rational expression that defines function F. Find the horizontal aymptotes for the resulting expression, then take the square root of those, to get from F^2 back to F.

Hint: There is a set of properties for rational functions (i.e., numerator and denominator both polynomials) that provide the horizontal asymptotes in terms of degrees, leading coefficients, and so forth. Have you seen these properties ? They offer an alternative to dividing top and bottom by the largest power of x.

I welcome specific questions.

 

[galactus]

To find of something like \(\displaystyle \lim_{x\to \infty}\frac{x-9}{\sqrt{4x^{2}+3x+2}}\)

you can divide the top by x and the bottom by \(\displaystyle \sqrt{x^{2}}\).

In doing so, we get:

\(\displaystyle \Large\lim_{x\to \infty}\frac{\frac{\not{x}^{1}}{\not{x}}-\frac{9}{x}}{\sqrt{\frac{4\not{x^{2}}}{\not{x^{2}}}+\frac{3\not{x}}{x^{\not{2}^{1}}}+\frac{2}{x^{2}}}}\)

As \(\displaystyle x\to {\infty}\), the terms with x in the denominator head toward 0 and we are left with........what?.

 

[jpnov]

^ wouldnt that leave a 0 in the numerator? So the horizontal asymptote is 0? Nevermind I see it's 1 on the top. Still working on it..

Thank you both for your help

 

[jpnov]

I got 1/2 as the horizontal asymptote. I divided everything by sqrt x^2, and after subbing in infinity, I got 1/(sqrt 4).

 

[mmm4444bot]

I got 1/2 as the horizontal asymptote.

y = 1/2 is one-half correct.

after subbing in infinity

This is not properly phrased because infinity is not a number; we cannot substitute infinity into arithmetic.

I get two horizontal asymptotes.

Remember: sqrt(x^2) does not equal x; it equals |x|.

When considering a function's global behavior, we must examine what's happening at both extremes of the domain. In exercise 34, that means considering what happens to F(x) as x approaches both positive and negative infinity.

Cheers

PS: What do you say about any vertical asymptotes ?

 

[lookagain]

i hope this helps with 24.

The limit of \(\displaystyle \frac{ax - bx}{\sqrt{ax + x^2} + \sqrt{bx + x^2}}\) as \(\displaystyle x\) approaches \(\displaystyle \infty\) is \(\displaystyle \frac{a - b}{2}:\)

There is not a justification for this step, and there must be at least one intermediate (connecting) step.

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\(\displaystyle \lim_{x \to \infty} \frac{\frac{1}{x}(ax - bx)}{\frac{1}{x}(\sqrt{ax + x^2} + \sqrt{bx + x^2})} \ \ =\)

\(\displaystyle \lim_{x \to \infty} \frac{a - b}{\sqrt{\frac{ax}{x^2} + \frac{x^2}{x^2}} + \sqrt{\frac{bx}{x^2} + \frac{x^2}{x^2}}} \ \ =\)

\(\displaystyle \lim_{x \to \infty} \frac{a - b}{\sqrt{\frac{a}{x} + 1} + \sqrt{\frac{b}{x} + 1}} \ \ =\)

\(\displaystyle \ \ \frac{a - b}{\sqrt{0 + 1} + \sqrt{0 + 1}} \ \ =\)

\(\displaystyle \ \ \frac{a - b}{1 + 1} \ \ =\)

\(\displaystyle \boxed{\frac{a - b}{2}}\)

 

[jpnov]

Thank you guys. That helped a lot, and most importantly I understand how to do the problems fully now