lim .............. 2^1/x =
x=>0-
2006-09-24 05:36:13 · 6 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
the limit is 0
the reason is that exponential functions are continuous so we can
take the limit of 1/x as x-->0-, in this case we get that 1/x gets to be more and more negative, but 2^{negative numbers} is small.
so the limit is =0
2006-09-24 05:38:45 · answer #1 · answered by Anonymous · 1⤊ 0⤋
If you're trying to ask about
lim ... 2^(1/x)
x->0-
that would be 2^(1/
which would be 2^(
which in turn is 0
2006-09-24 12:42:18 · answer #2 · answered by felix_doc 2 · 1⤊ 0⤋
lim(x>0-) 1/x = -infiniti
2^(-infiniti) = 0
2006-09-24 12:45:07 · answer #3 · answered by Anonymous · 1⤊ 0⤋
limit is 0
2006-09-24 13:16:33 · answer #4 · answered by m s 3 · 0⤊ 0⤋
substitute x=0-h
x->0 h->0 therefore
lim ---------- 2^1/-h
h->0
(1/2^inf)
=0 ans
2006-09-24 12:43:42 · answer #5 · answered by Gunjit M 2 · 1⤊ 0⤋
no
2006-09-24 12:44:22 · answer #6 · answered by Davette T 1 · 0⤊ 0⤋