Another Calculus Homework Problem I can not figure out. Can anyone solve it?

Question

If 1/x < or = f(x) < or = 1/x^3 (which it is if x < -1)

then lim f(x) =
.... . x=>-oo

Answer 1

No calculus here ... only a limits problem.

The answer is, f(x) approaches zero from the negative side as x approachs negative infinity.

In this problem, you don't have to worry about either positive or small values of x. The only thing that matters is large negative x.

Here's two ways to see it. Pick some large negative number such as (-100). Then

1. 1/(-100) <= f(x) <= 1/(-100)^3
-0.01 <= f(x) <= -0.000001

Here, you can see that f(x) is always negative, is always between the limits, and, as x gets larger in the negative direction, the limits get closer to zero.

2. Multiply that inequality by (-1) (to make it positive), and of the inequality. Then you have:

-1/(x) >= -f(x) >= -1/(x^3)
-1/x <= -f(x) >= -1/x^3

Pick a large negative number such as -100. Then

-1/(-100) >= -f(x) >= -1/(-100)^3
1/100 >= -f(x) >= 1/1000000
0.01 >= -f(x) >= 0.000001

Here, looking at it from the positive side, the limits again approach zero from the positive side, so [-f(x)] also approaches zero from the positive side; but that means f(x) (without the negative sign) approaches it from the negative side.

Often this limit is written as "0-" ... in other words, you could write lim f(x) = 0- etc ...

Answer 2

The answer is zero. If you plug in negative infinity for "x" in the given system of inequalities, you get
"-0 is < or = to f(-oo) which is < or = -0" .

I'm not "in" calculus yet per se, but I'm just exceptional at mathematics, and I've studied a bit of calculus in my day ;).

(P.S. You've probably learned this, but:
The lim as x=>+/-oo of f(x)= 1/x is = to 0

and
(oo)^|n|, where n is any real, non-zero number is = to (oo).)

Answer 3

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Answer 4

f(x)=0 ... 1/oo = 0 and 1/(oo)^3 = 0

Answer 5

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