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.


Evaluate $$L=\lim_{x\to0} {\frac{\left(1+\sin{x}+\sin^2{x}\right)^{1/x}-\left(1+\sin{x}\right)^{1/x}}{x}}$$

If there are more terms to apply the limit to, you should separate the terms of which limits are apparent to reduce the probability of such failure.

\begin{align}
L&=\lim_{x\to0} {\frac{\left(1+\sin{x}+\sin^2{x}\right)^{1/x}-\left(1+\sin{x}\right)^{1/x}}{x}}\\
&=\lim_{x\to0} \left(1+\sin{x}\right)^{1/x}{\frac{\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)^{1/x}-1}{x}}\tag{1}\\
&=e\lim_{x\to0} {\frac{\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)^{1/x}-1}{x}}\tag{2}\\
&=e\lim_{x\to0} {\frac{\exp\left[\frac 1x\ln\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)\right]-1}{x}}\tag{3}\\
&=e\lim_{x\to0} {\frac{\exp\left\{\displaystyle\frac 1x\left[\sum_{i=1}^{+\infty}(-1)^{i+1}\frac1i\left(\frac{\sin^2{x}}{1+\sin{x}}\right)^i\right]\right\}-1}{x}}\tag{4}\\
&=e\lim_{x\to0} {\frac{{\displaystyle\sum_{j=1}^{+\infty}\frac{\left\{\displaystyle\frac 1x\left[\sum_{i=1}^{+\infty}(-1)^{i+1}\frac1i\left(\frac{\sin^2{x}}{1+\sin{x}}\right)^i\right]\right\}^j}{j!}}}{x}}\tag{5}\\
&=e\lim_{x\to0} {\frac{\frac 1x\frac{\sin^2{x}}{1+\sin{x}}}{x}}=e\tag{6}\\
\end{align}
where in step $(6)$, I used the term $i=j=1$ only, because

*

*$\displaystyle\lim_{x\to 0}(1+\sin x)=1$.

*$\displaystyle\lim_{x\to 0}\frac{\sin^a x}{x^b}=1\Leftrightarrow a=b$, where $a,b\in\mathbb R$.

*$a\rightarrow 2ij=b\rightarrow j+1\Rightarrow j(2i-1)=1$, where $i,\ j\geq 1$

You did something like $\displaystyle\lim_{x\to 0}\frac{e-e}{0}$ in the place of my step $(4)$ because of overwhelmingly many terms on both sides of the minus sign.
