dilemma help?

Question

Find

lim x ->0 (x^0)

and

lim x->0 (0^x)

and

lim x->0 (x^x)

And please explain why the values are not the same.

^_^

PLEASE EXPLAIN why the values are not the same.

Answer 1

The 1st limit:
You should take in to the account that the x tends to 0, meaning it is not actually 0, and therefore you r function is x^0 equals, by the definition to 1. Hence, the limit equals to 1 (as it is the limit of a sequence of 1's).

The 2nd limit:
Here, first of all, should be mentioned that the x tends to 0+, otherwise you end up with a division by 0. Secondly, again by the definition, 0 in any positive power is 0. Hence, your limit equals to 0.

The 3rd limit:
It is a special case of lmit of function briefed as "0^0". It is treaded as following (using the the L'Hopital's rule):
x^x=Exp(x*ln x)=Exp(ln x/(1/x)). Therefore, the original limit is the limit of the exponent in the power of the limit of ln x/(1/x) (if the latter exists and the function, and in this particular case it is, is continuous). To calculate the limit of ln x/(1/x) use the L'Hopital's rule and find the limit of the quotient of the derivatives of the nominator and denominator. You'll see it is 0, and therefore the value of the original limit is Exp (0)=1.

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Answer 2

Since anything to the 0 power except 0^0 = 1, that limit is 1.

Since 0 to any power except 0 is 0, that limit is 0

Since you can't do a negative decimal to a negative decimal, the third one only exists on one side of 0.

Answer 3

a way of approaching the limit of a number is to make X a number close enough to the limit wanted. Try pluging in your calculator a number so small that could be close enough to 0.
For example, for the lim x->0 (x^0) try (1E-20)^0
1E-20 is so small than is close ebough to 0. Find the results by yourself

Answer 4

http://en.wikipedia.org/wiki/Empty_product

if you consider the infinite taylor series(convenient around zero) for the exponential function exp z or e^z
the first term is 0^0 / 0!
the denominator can be shown to be 1 and we know that for any number to the zeroth power is unity
z^0 =1
so 0^0 is often unity. However when given as variables factors such as approach come into effect and the exponential may or may not converge.(that answers your last limit)
0^x is often considered zero as well so you might be better off going with undefined.

Answer 5

x^0 = 1 since x can never = 0, only approach it

0^x = 0 since x can never = 0, only approach it

x^x is undefined since 0^0 is undefined