Question 847244
Limx>0 (1-2x/1+2x)^1/3x
<pre>
You didn't put parentheses to show where your numerators and
denominators begin and end, so I will just guess that you meant
this:

{{{matrix(3,2,
"", "",
lim, ((1-2x)/(1+2x))^(expr(1/3)x),
"x->0", "")}}}

We find the limit of the natural logarithm
of the expression

{{{matrix(3,2,
"", "",
lim, ln(((1-2x)/(1+2x))^(expr(1/3)x)),
"x->0", "")}}}

Use a rule of logarithms:

{{{matrix(3,2,
"", "",
lim, ((expr(1/3)x)ln((1-2x)/(1+2x))^""),
"x->0", "")}}}

When we substitute 0 for x into

{{{(expr(1/3)x)ln((1-2x)/(1+2x))^""}}}

{{{(expr(1/3)0)ln((1-2*0)/(1+2*0))^""}}}

{{{(0)ln(1/1)}}} = 0

Since the natural logarithm of the function approaches 0,
the function must approach the number which has the natural
logarithm of 0, which is 1.  Therefore we can conclude:

{{{matrix(3,2,
"", "",
lim, ((1-2x)/(1+2x))^(expr(1/3)x),
"x->0", "")}}}{{{""=""}}}{{{1}}}

Edwin</pre>