a different problem was recently posted here where the left limit is 2 and the right limit is zero. Tweaking it a bit, we have
(8)/(1-2^(-1/x))
which fits your criteria.
If you can come up with any function where the left and right limits for some value of x are different, you can tweak it into another one which fits the criteria.
Create two different functions for which the left-hand limit at zero is 0 and for which the right hand limit at zero is 8
6 answers
I still don't understand
lim X->0= (0)/(1+8^(1/x)
as x-> 0, 1/x -> ∞
2^(-1/x) -> 2^-∞ = 0 when x>0
2^(-1/x) -> 2^∞ = ∞ when x<0
So, in those two cases, we have
lim x->0+ = 8/(1-2^-∞) = 8/(1-0) = 8
lim x->0- = 8/(1-2^∞) = 8/-∞ = 0
http://www.wolframalpha.com/input/?i=%288%29%2F%281-2^%28-1%2Fx%29%29
2^(-1/x) -> 2^-∞ = 0 when x>0
2^(-1/x) -> 2^∞ = ∞ when x<0
So, in those two cases, we have
lim x->0+ = 8/(1-2^-∞) = 8/(1-0) = 8
lim x->0- = 8/(1-2^∞) = 8/-∞ = 0
http://www.wolframalpha.com/input/?i=%288%29%2F%281-2^%28-1%2Fx%29%29
I am sorry i typed the problem wrong. It says Create two different functions for which the left-hand limit at zero is 0 and for which the right hand limit at zero is negative 8
well, since -0 = 0 and -8 = -(8), it should not be hard to make the required change to the function I provided.
I mean, you do remember your basic algebra, right?
I mean, you do remember your basic algebra, right?