Question 890059
<pre>
(1)  {{{matrix(2,1,"",lim["x->0"](((tanx)/x)^(1/x^2)))}}}

We will first find the limit of the natural log of the expression
and then the above will be e raised to the power of the result.

(2)  {{{matrix(2,1,"",lim["x->0"](ln((tanx)/x)^(1/x^2)))}}}{{{""=""}}}{{{matrix(2,1,"",lim["x->0"](expr(1/x^2)ln((tanx)/x)))}}}{{{""=""}}}

{{{lim["x->0"]  (   ln(tan(x)/x) /x^2)  }}}{{{""=""}}}

Since (A)  {{{lim["x->0"](tan(x)/x)=1}}} and {{{ln(1)=0}}}, and
      (B)  {{{lim["x->0"](x^2)=0}}}

we can use L'Hopital's rule since it has the form {{{0/0}}}.

We need to take the derivative of the numerator and the denominator:

Derivative of numerator:

     {{{matrix(1,2,d/(dx),(ln(tan(x)/x)))}}}{{{""=""}}}{{{matrix(1,2,d/(dx),( 

ln(tan(x))-ln(x)

)) }}}{{{""=""}}}{{{sec^2x/tan(x)-1/x}}}{{{""=""}}}
     {{{(1+tan^2(x))/tan(x)-1/x}}}{{{""=""}}}{{{1/tan(x)+tan^2(x)/tan(x)-1/x}}}{{{""=""}}}{{{cot(x)+tan(x)-1/x}}}{{{""=""}}}
     {{{cos(x)/sin(x)+sin(x)/cos(x)-1/x}}}{{{""=""}}}{{{(x*cos^2(x)+x*sin^2(x)-sin(x)cos(x) )/(x*sin(x)cos(x))}}}{{{""=""}}}
     {{{(x(cos^2(x)+sin^2(x))-sin(x)cos(x) )/(x*sin(x)cos(x))}}}{{{""=""}}}{{{(x(1)-sin(x)cos(x) )/(x*sin(x)cos(x))}}}{{{""=""}}}
     {{{(x-sin(x)cos(x) )/(x*sin(x)cos(x))}}}{{{""=""}}}

     (multiply by {{{2/2}}} to make use of identity sin(2x)=2sin(x)cos(x):
     {{{(2x-2sin(x)cos(x) )/(x*2sin(x)cos(x))}}}{{{""=""}}}{{{(2x-sin(2x) )/(x*sin(2x))}}}

Derivative of denominator:  {{{matrix(1,2,d/(dx),(x^2))}}}{{{""=""}}}{{{2x}}}

So (2) becomes:

{{{lim["x->0"](expr(expr(( ( 2x-sin(2x) )/(x*sin(2x)    ) ))/(2x)) ) )}}}{{{""=""}}}{{{lim["x->0"]((  2x-sin(2x) )/(2x^2*sin(2x)     ) ) )}}}{{{""=""}}}

Use L'Hopital's rule again:

{{{lim["x->0"]((2-2cos(2x))/(4x^2cos(2x)+4x*sin(2x)))}}} (divide top and bottom by 2)


{{{lim["x->0"](

(1-cos(2x))/(2x^2cos(2x)+2x*sin(2x)))}}} 

Use L'Hopital's rule again:

{{{lim["x->0"](

2sin(2x)/(-4x^2sin(2x)+4x*cos(2x)+4x*cos(2x)+2sin(2x)))}}}

{{{lim["x->0"](

2sin(2x)/(-4x^2sin(2x)+8x*cos(2x)+2sin(2x)))}}}

Divide top and bottom by 2

{{{lim["x->0"](
sin(2x)/(-2x^2sin(2x)+4x*cos(2x)+1sin(2x)))}}}

Use L'Hopital's rule once more:

{{{lim["x->0"](

2cos(2x)/(-4x^2cos(2x)-4x*sin(2x)-8x*sin(2x)+4cos(2x)+2cos(2x)))}}}{{{""=""}}}

No don't need to bother simplifying that, because we can just substitute
x=0

{{{2/(-0-0-0+4+2)}}}{{{""=""}}}{{{2/6}}}{{{""=""}}}{{{1/3}}}

Therefore the log of the answer is {{{1/3}}}, which means:

{{{matrix(2,1,"",lim["x->0"](((tanx)/x)^(1/x^2)))}}}{{{""=""}}}{{{matrix(2,3,
"",      "",      "",
e^(1/3),""="",root(3,e))}}}




Edwin</pre>