Calculate the limit if it exists.

I wasn't sure if to post this in the Calculus forum or in the Algebra forum. My question has more to do with the algebra than it does with the calculus part however since it is a limit question I thought it would be better to post to the Calculus forum.

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It looks to me by looking at the graph that as x->infinity that y-> -infinity. Problem f) seems to be solved positive infinity? I'm still trying to figure out the algebra that I highlighted I don't see yet how those are equivalent. Also I was wondering am I supposed to know the answer to f) without looking at its graph or how if that is possible?

Thanks in advance...

type out your work and then I can help you find the error

on a side note - learning to use LaTeX will greatly aid in typing out your work. I find that because it takes more time than writing it out on paper you get a better understanding of a problem you are having problems with. It is fairly easy to learn.

You have to multiply the whole fraction by $\displaystyle \frac { \frac { 1 }{ x } }{ \frac { 1 }{ x } } $ wich is secretly 1 so you are not changing the function.
$\displaystyle \frac { 4{ x }^{ 2 } }{ x-2 } \ast \frac { \frac { 1 }{ x } }{ \frac { 1 }{ x } } =\frac { \frac { 4{ x }^{ 2 } }{ x } }{ \frac { x-2 }{ x } } =\frac { 4x }{ \frac { x }{ x } -\frac { 2 }{ x } } =\frac { 4x }{ 1-\frac { 2 }{ x } } $ so we say both fractions are equivalent and the second form is useful to find the limit.

I hope this is useful

PD: You could also use L'Hôpital Rule to find this limit

Quote:

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Originally Posted by ManuelSG

You have to multiply the whole fraction by $\displaystyle \frac { \frac { 1 }{ x } }{ \frac { 1 }{ x } } $ wich is secretly 1 so you are not changing the function.

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Why do you say that? There is nothing secret about that.

But why not just say "divide the numerator and the denominator by x"?

Oh another trick in your book is looking at the degree of the dominant term in the denominator and numerator.

And,

If the degree in the dominant term of both the denominator and numerator are equal you'll end up with a horizontal asymptote at y=0

or

If the degree of the dominant term in the denominator is greater the the degree of the dominant term in the numerator you will have a horizontal asymptote

at $\displaystyle y = \frac{coefficient\, of\, the\, dominant\, term\, in\, the\, numerator}{coefficient\, of\, the\, dominant\, term\, in\, the\, denominator}$

but in yours the greater degree of a dominant term is in your numerator thus $\displaystyle f(x) \rightarrow \pm \infty \,as\,x\rightarrow \pm \infty$

And then you can test the behavior of the function by creating the quotient $\displaystyle \frac{dominant\, term\, in\, the\, numerator}{dominant\, term\, in\, the\, denominator}$

and see how it behaves when x gets large and negative and large and positive


Also,

it might help you to remember

$\displaystyle \lim_{x \ \pm\infty} \frac{1}{x} = 0$ so the idea is to express $\displaystyle f(x) $ in terms of $\displaystyle \frac{1}{x}$

hope those help!

To begin with thank you for all your very helpful replies!

$\displaystyle \frac{4x}{1-\frac{2}{x}}$

I see now how the above equation was derived. Also thank you Jonroberts74 for posting:

$\displaystyle \lim_{x\pm\infty}\frac{1}{x}=0$
The above is a concept that I can understand.

In my solutions manual there is an equation:

$\displaystyle \frac{4x}{1-\frac{2}{x}}\rightarrow \frac{\infty}{1}$

I have to admit I have been looking at it for quite some time and I don't see yet what is meant by:

$\displaystyle {\rightarrow \frac{\infty}{1}}$

Also as I have graphed the equation which is attached to my original post and by looking only at the graph there is no $\displaystyle f(x)$ where x is positive that yields a positive y value so I still don't see how the limit of the function is positive infinity as x goes in the positive direction towards infinity.

I had to fix some Latex errors. My tag should have been [tex] not [math]

$\displaystyle \frac{\infty}{1} = \infty\frac{1}{1} = \infty$

thus $\displaystyle as\, f(x) \rightarrow \infty$

also sorry, it should be

$\displaystyle \lim_{x\rightarrow\pm\infty} \frac{1}{x} = 0$ and not $\displaystyle \lim_{x\pm\infty} \frac{1}{x} = 0$ I forgot the $\displaystyle \rightarrow$

$\displaystyle \frac{4\infty}{1-\frac{2}{\infty}}\rightarrow \frac{\infty}{1-0}$

So could I say that the equation of the left and the equation on the right are equivalent?

Quote:

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Originally Posted by sepoto

$\displaystyle \frac{4\infty}{1-\frac{2}{\infty}}\rightarrow \frac{\infty}{1-0}$

So could I say that the equation of the left and the equation on the right are equivalent?

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Well this might be getting into some territory that I don't quite know yet but to me it seems like you are now treating infinity as though it is a number or variable which is incorrect.

Regardless,

I would state it as "f(x) tends to infinity".

It looks like I should have zoomed out quite a lot with my graph.Attachment 29626