Limits

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July 27th 2006, 11:51 PM
askmemath

Limits

Consider f (x) = (-7x^3 + 2)/(10x^3 + 3x + 2)

a) Construct a table of values to determing lim f(x) x to infinity

b) Use algebra to determin lim f(x)
x to infinity

c) Sketch the graph and indicate on the graph the behavior described by the limit.
July 28th 2006, 03:58 AM
galactus

part a. plug and chug.

For part b, get rid of everything but the highest powers in the num. and den.

You are left with: $\frac{-7x^{3}}{10x^{3}}$

Now, you can see what the limit is, can't you?.

part c. You have a graphing calculator?.
July 28th 2006, 07:06 AM
ThePerfectHacker

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July 28th 2006, 08:50 AM
askmemath

Quote:

Originally Posted by galactus

part a. plug and chug.

For part b, get rid of everything but the highest powers in the num. and den.

You are left with: $\frac{-7x^{3}}{10x^{3}}$

Now, you can see what the limit is, can't you?.

part c. You have a graphing calculator?.

Umm am not really sure I understand what you meant y plug and chug.
Can I plug in any value that I like? Arent I supposed to follow a sequence or something??

For part B how can I just get rid of everything but the highest powers?
L'Hospital Rule??
July 28th 2006, 09:15 AM
galactus

Just plug in higher and higher values and watch it approach the limit.

$\frac{-7(1)^{3}+2}{10(1)^{3}+3(1)+2}=\frac{-1}{3}$

$\frac{-7(100)^3+2}{10(100)^{3}+3(100)+2}=\frac{-3499999}{5000151}=-.699978660644$

$\frac{-7(1000)^{3}+2}{10(1000)^{3}+3(1000)+2}=\frac{-3499999999}{5000001501}=-.69999978966$

See, it's approaching $-.7\;\ or\;\ \frac{-7}{10}$

Now, make it official with part b. Rational functions, especially with polynomials, are one of the easiest things in which to take a limit.

For part b, no, just erase them. In a rational function, if the limit approaches infinity, it is unaffected if you discard everything but the highest powers. Think about it. See?.

Suppose I had $\lim_{x\to\infty}\frac{3x+5}{6x-8}$

If x gets huge, towards infinity, what are the 5 and 6 going to amount to?. Nothing. So, get rid of them and take the limit of 3x/6x. Which is 3/6=1/2.

Yours works the same.

$\frac{-7x^{3}}{10x^{3}}$
July 28th 2006, 09:31 AM
ThePerfectHacker

Quote:

Originally Posted by askmemath

For part B how can I just get rid of everything but the highest powers?
L'Hospital Rule??

You have,
$\frac{-7x^3 + 2}{10x^3 + 3x + 2}$
Divide numerator and denominator by $x^3$ ,
$\frac{-7+\frac{2}{x^3}}{10+\frac{3}{x^2}+\frac{2}{x^3}} \to -\frac{7}{10}$ as, $x\to\infty$
Becuase,
$\frac{k}{x^n}\to 0$