Question

CALCULUS HELP: Consider the function f(x) = 100 - (12000)/(400+3000a^x), where 0 < a < 1. Which of the following are true?

Answer 1

There can be more than one answer:
I. The limit of f(x) = 70 as x reaches infinity.
II. The limit of f(x) = 70 as x reaches negative infinity.
III. The limit of f(x) = 100 as x reaches infinity.
IV. The limit of f(x) = 100 as x reaches negative infinity.
V. The limit of f(x) =100 - (12000/400+3000a) as x reaches 0.

Answer 2

\[\Large\bf\sf f(x)\quad=\quad 100-\frac{12000}{400+3000a^x}\]
\[\Large\bf\sf \lim_{x\to\infty}f(x)\quad=\quad 100-\frac{12000}{400+3000(0)}\]Since x is getting BIGGGGG,
and our a is a fraction beetween 0 and 1,
it means the a^x is getting really small,
approaching zero.

Is the first one true or no?

Answer 3

Or I guess I should say, use this information to determine if 1 or 3 is correct.
They can't both be true.

Answer 4

Negative infinity is a little trickier, try to think about your rules of exponents.
Here I'll give an example:

if a = 1/10,
and x is really really large, we might have something like this,\[\Large\bf\sf \left(\frac{1}{10}\right)^{-99999}\quad=\quad \left(\frac{10}{1}\right)^{99999}\quad=\quad 1\;bajillion\]

Answer 5

See how the a^x blows up really huge as x -> -infty ?

Answer 6

Oh, sorry! Didn't realize you were typing lol. Okay, give me a moment to process what you just said.

Answer 7

XD

Answer 8

@zepdrix Okay, I'm going to go on a limb here and say that the first one is true. As x reaches infinity, the limit of the equation is 70. Is my thinking correct here? ,__, So far, the correct answers would be I and VI, right?

Answer 9

So as x-> infnty,\[\Large\bf\sf \lim_{x\to\infty}f(x)\quad=\quad 100-\frac{12000}{400}\quad=\quad 70\]Mmmm ya that one looks good!

Answer 10

And as x-> -infty,\[\Large\bf\sf \lim_{x\to-\infty}f(x)\quad=\quad 100-\frac{12000}{1\;bajillion}\quad=\quad 100-0\quad=\quad 100\]
Yayyy good job!
I think you meant to type IV, instead of VI, right? :P heh

Answer 11

@zepdrix Oh yes! Lol. Just to be sure, V is also true too right? So it would be I, IV and V.

Answer 12

There is no discontinuity or anything fancy going on here,
\[\Large\bf\sf \lim_{x\to0}a^x\quad=\quad a^0\]We can plug the zero directly in.
It's approaching a^0, which is just 1, right?

Answer 13

Ah I see. Thanks for helping!