<s>[INST] 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.
 [/INST] Hint: Write the function of interest as
$$
(1+\sin x)^\frac{1}{x}\,\frac{e^{\frac{1}{x}\log(1+\frac{\sin^2 x}{1+\sin x})}-1}{x}
$$
and remember that
$$
\lim_{x\to0}\frac{\sin x}{x}=1\,,\qquad \lim_{x\to0}\frac{\log(1+x)}{x}=1\,,\qquad \lim_{x\to0}\frac{e^{x}-1}{x}=1\,.
$$
You should find that the limit is $e$.
</s>