Takehome Calculus Quiz - Topic

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Sep 17 2013 06:52pm

Quote (brmv @ Sep 17 2013 08:39pm)

if you are looking for the limits when x goes to +inf and -inf then it is just like question 3, see at 'Azrad's explanation above (assuming you take the positive square root)
sqrt(x^2-4) ~ absolute value of x, ie you get (3x+2)/x now you can forget the 2 for very large x and you have 3x/abs(x)

and if you do not know what a complex number is, forget atm all about it

looking at the denominator:

i see how sqrt(x^2-4) is simplified to
sqrt(x^2)
x

you're saying that x is an absolute value?
and that leads you to plug in both a + and - (so that the two answers are 3,-3?)

after lookin at the problem for a while i realized i could simplify it to 3, but im confused as to where the -3 comes into play

brmv

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#22

Sep 17 2013 06:56pm

Quote (known954 @ 18 Sep 2013 00:52)

looking at the denominator:
i see how sqrt(x^2-4) is simplified to
sqrt(x^2)
x
you're saying that x is an absolute value?
and that leads you to plug in both a + and - (so that the two answers are 3,-3?)
after lookin at the problem for a while i realized i could simplify it to 3, but im confused as to where the -3 comes into play

sqrt(x^2) does have 2 values, namely -x and +x [the absolute value of a number is the positive component abs(-x)=abs(x)]

now if you take the positive sqrt of x when you go to -inf it will be the -3
[likewise if you take the negative square root when you go to +inf,
basically you have 2 functions because the square root has two values]
you understand now?

This post was edited by brmv on Sep 17 2013 07:02pm

known954

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#23

Sep 17 2013 07:13pm

i think so, if i have any more questions ill be sure to ask.

thanks again! really appreciate the help

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