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.

Your second last equality is not justified: You cannot take limit of enumerator and denominator separately as the limit of the denominator is (trivially) equal to 0. Try the appropriate version of l'Hospitals rules (which helps, because enumerator and denominator (as you have formally shown) converge to $0$.
