Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
 \frac
 {
  \left(
   1+\sin{x}+\sin^2{x} 
  \right)
  ^{1/x}
  -
  \left(
   1+\sin{x}
  \right)
  ^{1/x}
 }
 {
  x
 }
}
$$
Let's begin then.
Using $f(x) = e^{\log{f(x)}} = \exp{\left[\log{f(x)}\right]}$ it gets to:
$$
\lim_{x\to0} {
 \frac
 {
  \exp
   \left[
   \frac
   {
    \log{\left(1+\sin{x}+\sin^2{x}\right)}
   }
   {
    x
   }
   \right]
  -
  \exp
   \left[
   \frac
   {
    \log{\left(1+\sin{x}\right)}
   }
   {
    x
   }
   \right]
 }
 {
  x
 }
}
$$
Multiplying and dividing by the right terms:
$$
\lim_{x\to0} {
 \frac
 {
  \exp
   \left[
   \frac
   {
    \log{\left(1+\sin{x}+\sin^2{x}\right)}
   }
   {
    \sin{x}+\sin^2{x}
   }
   \frac
   {
    \sin{x}+\sin^2{x}
   }
   {
    x
   }
   \right]
  -
  \exp
   \left[
   \frac
   {
    \log{\left(1+\sin{x}\right)}
   }
   {
    \sin{x}
   }
   \frac
   {
    \sin{x}
   }
   {
    x
   }
   \right]
 }
 {
  x
 }
}
$$
Grouping the $\sin{x}$ in the first exponential function:
$$
\lim_{x\to0} {
 \frac
 {
  \exp
   \left[
   \frac
   {
    \log{\left(1+\sin{x}+\sin^2{x}\right)}
   }
   {
    \sin{x}+\sin^2{x}
   }
   \frac
   {
    \sin{x}
   }
   {
    x
   }
   \left(1+\sin{x}\right)
   \right]
  -
  \exp
   \left[
   \frac
   {
    \log{\left(1+\sin{x}\right)}
   }
   {
    \sin{x}
   }
   \frac
   {
    \sin{x}
   }
   {
    x
   }
   \right]
 }
 {
  x
 }
}
$$
Applying the know limits, specifically:
$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}+\sin^2{x}\right)}}{\sin{x}+\sin^2{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}\right)}}{\sin{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\sin{x}}{x}=1}$$$$\lim \limits_{x\to0}{\left(1+\sin{x}\right)}=1$$
We should end up with:
$$
\lim_{x\to0} {
 \frac
 {
  \exp
   \left[
   \frac
   {
    \log{\left(1+\sin{x}+\sin^2{x}\right)}
   }
   {
    \sin{x}+\sin^2{x}
   }
   \frac
   {
    \sin{x}
   }
   {
    x
   }
   \left(1+\sin{x}\right)
   \right]
  -
  \exp
   \left[
   \frac
   {
    \log{\left(1+\sin{x}\right)}
   }
   {
    \sin{x}
   }
   \frac
   {
    \sin{x}
   }
   {
    x
   }
   \right]
 }
 {
  x
 }
}
=
\frac
{
 \exp{\left[1(1)(1)\right]}
 -
 \exp{\left[1(1)\right]}
}
{
 0
}
 = \frac{e-e}{0}
 = \frac{0}{0}
$$
Undetermined form, yay. Any hint is really appreciated, and I'm really really sorry for all the spaghetti rendering, it's hard to look at.

You should note that the expression in numerator is of the form $A-B$ with both $A, B$ tending to $e$. We can write $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A-\log B} \cdot(\log A - \log B) $$ Since the middle factor tends to $1$ the desired limit is equal to the limit of $$e\cdot\frac{\log A-\log B} {x} =e\cdot\frac{\log(1+\sin x+\sin^2x)-\log(1+\sin x)} {\sin^2x}\cdot\frac{\sin^2x}{x^2}$$ The last factor tends to $1$ and putting $t=\sin x$ we see that desired limit equals the limit of $$e\cdot\frac{\log(1+t+t^2)-\log(1+t)}{t^2}=e\cdot\frac{\log(1+(t^2/(1+t)))}{t^2/(1+t)}\cdot\frac{1}{1+t}$$ and clearly the above tends to $e\cdot 1\cdot 1=e$.
