This limit function problem below is impossoble. Need your Help!?

Question

. . = x^2 + 3, if x < 0
. . = 0, if x = 0
f(x) = x^2 - 3 if 0 < x < 2
. . = 1 if x = 2
. . = x^3 - 7 if 2 < x

Then lim f(x) =
. .. . . x=>0+

Answer 1

(I take it the '...' don't mean anything has been truncated.)

So f(x) is just a piecewise-define function.

The piece relevant to determining lim x→0+ f(x) is:
f(x) = x^2 - 3 if 0 < x < 2

Then lim x→0+ f(x) = lim x→0+ x^2 - 3
= -3

(Obviously f(x) has a discontinuity at 0, since lim x→0+ f(x) = +3 , if that is what is confusing you. They never said it was continuous.)

Answer 2

The notation is unclear (hand to type math symbols in yahoo).

Does x=>0+ mean "as x approaches zero from the right"?
If it does, the x is in the range 0 < x < 2, so f(x) = x^2 -3.
Then lim f(x) is just 0^2 - 3 = 0 - 3 = -3.

But if x=>0+ means "as x approaches positive infinity", then that's another way of saying, "as x gets bigger and bigger ...".
Then for large x, we have f(x) = x^3 -7.
As x gets very large, the "-7" part gets less important.
So in the limit, we have f(x) --> x^3.

Answer 3

I don't remember how to solve a limit problem but the answer is obvious if you understand the question. The question is what is the limit, as x approaches zero (from the positive side), of x^2-3. (Doesn't matter that f(x) is defined as = 0 at zero, that is not the limit because we could just as easily say f(0)=43 or f(0)= -3.14159.. its arbitary and the limit of x^2-3 wouldn't change every time we decided to DEFINE f(0) differently - we can define it as - i if we want!
ANyway x squared approaches zero much faster than x does as x approaches zero ( if x= 0.1, x^2 = .01, , x=.001, x^2=.000001) so we know that the limit of x^2-3 as x approaches zero is -3. Now you do the calculus.