limits, asymptotes and studying the behaviour of f(x)

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dangerous_dave

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Mar 13, 2008
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I have a function
f(x)=
4x(x-2)
(x-1)^2

I have done most of the questions relating to it, but I'm having trouble with these 3:

1. Evaluate lim x-> infinity and negative infinity, stating whether f(x) approaches the limit from above or below. Justify your answers.

2. Write down the equations of all asymtotes.

3. Study the behaviour of f(x) as x->1+ and as x-> 1-

For 2, I think I have have to differentiate then solve for 0? The others I have no idea what is being asked.
 
You do not have to differentiate. The vertical asymptotes occur where you have division by 0.

The horizontal occurs where the limit x-->infinity. Check at \(\displaystyle {-\infty}\), also. The limit should be the same.

\(\displaystyle \lim_{x\to\infty}\frac{4x(x-2)}{(x-1)^{2}}\)

\(\displaystyle =\lim_{x\to\infty}\left[4-\frac{4}{(x-1)^{2}}\right]\)

See what the limit is and, therefore, the horizontal asymptote?.

As far as the limit approaching 1 from the left or right, you should be able to see what that would be considering the x-1 in the denominator.
 
So the vertical is at x=1?

I thnk you rearanged the equation, but I don't see how. Is it a horozontal asymptote ate x=2?
 
In addition - plot the function using your calculator (or computer) and see those asymptotes.
 
Sorry, its 4, right?
Im having trouble with the long division, do I have these round the right way?

..............________
x^2-2x+1)4x-8

I can only find examples for when the x powers are greater for the top of the equation,

I have plotted in on a calculator. But im not (much) interested in the final result, but more how to get there.
 
dangerous_dave said:
Sorry, its 4, right?
Im having trouble with the long division, do I have these round the right way?

..............________
x^2-2x+1)4x^2 - 8x


The way I would do is as follows:

\(\displaystyle \frac{4x^2-8x}{x^2 - 2x + 1}\)

\(\displaystyle =\, \frac{4\cdot (x^2-2x+1) - 4}{x^2 - 2x + 1}\)

\(\displaystyle =\, 4 \, -\, \frac{4}{x^2 - 2x + 1}\)

I can only find examples for when the x powers are greater for the top of the equation,

I have plotted in on a calculator. But im not (much) interested in the final result, but more how to get there.

Plotting is advisable not just for answer. It would show that the asymptote is at 4 - and spare you the wasted effort and time for asking whether it is 2.
Click to expand...
 
dangerous dave: For what it is worth Dept. If you have a TI-89, after doing all your grunt work manually,

as Subhotosh Klan stated, check your work with a calculator, as this is easily done.

This way any errors can be detected (and corrected) giving you an A+ on your homework instead of the infamous big fat

goose egg.
 
galactus said:
You do not have to differentiate. The vertical asymptotes occur where you have division by 0.

The horizontal occurs where the limit x-->infinity. Check at \(\displaystyle {-\infty}\), also. The limit should be the same.

>> WELL, NOT ALWAYS. {In this case, yes.]



\(\displaystyle \lim_{x\to\infty}\frac{4x(x-2)}{(x-1)^{2}}\)

\(\displaystyle =\lim_{x\to\infty}\left[4-\frac{4}{(x-1)^{2}}\right]\)

See what the limit is and, therefore, the horizontal asymptote?.

As far as the limit approaching 1 from the left or right, you should be able to see what that would be considering the x-1 in the denominator.
Click to expand...
 
Why worry about the long division. You probably should do that for the practice, but it's a bit excessive, here. Just look at it.

Degree of Numerator < Degree of Denominator -- Horizontal Asymptote is y = 0

Degree of Numerator = Degree of Denominator -- Horizontal Asymptote is y = a/b, where a is the coefficient of the term of highest degree in the numerator and b is the coefficient of the term of highest degree in the denominator.

Degree of Numerator > Degree of Denominator -- Okay, long division may be the way to go on this one.
 
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