What is the following limit? How can I calculate the following limit?
What do I miss here? What am I doing wrong?
$$\begin{align*}
\lim_{x\rightarrow0}\frac{6x\cdot\sin x-6\cdot\sin\left(x^{2}\right)+x^{4}}{x^{5}\left(e^{x}-1\right)}= & \lim_{x\rightarrow0}\frac{6x\cdot\sin x}{x^{5}\left(e^{x}-1\right)}-\lim_{x\rightarrow0}\frac{6\cdot\sin\left(x^{2}\right)}{x^{5}\left(e^{x}-1\right)}+\lim_{x\rightarrow0}\frac{x^{4}}{x^{5}\left(e^{x}-1\right)}=\\
= & \lim_{x\rightarrow0}\frac{6}{x^{3}\left(e^{x}-1\right)}\cdot\underbrace{\frac{\sin x}{x}}_{\rightarrow1}-\lim_{x\rightarrow0}\frac{6}{x^{3}\left(e^{x}-1\right)}\cdot\underbrace{\frac{\sin\left(x^{2}\right)}{x^{2}}}_{\rightarrow1}+\lim_{x\rightarrow0}\frac{1}{x\left(e^{x}-1\right)}=\\
= & \lim_{x\rightarrow0}\frac{1}{x\left(e^{x}-1\right)}=\infty\neq\frac{21}{20}.
\end{align*}$$
According to WolframAlpha $21/20$ is the solution. What am I doing wrong?
 A: You replaced parts of the formula with their limits, so that some terms (not replaced with their limit) cancel. This may lead to wrong results, as you see.
I suggest finding equivalents for the numerator and the denominator.

*

*For the denominator, it is basid:  it is known from Taylor-Young's formula that $\mathrm e^x-1\sim_0 x$, hence $\;x^5(\mathrm e^x-1)\sim_0x^6$

*For the numerator, apply Taylor-Young's formula to each term so as to obtain ultimately an expansion at order $6$:
\begin{align}
6x\sin x-6\sin x^{2}+x^4 &= 6x\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)-\Bigl(6x^2-\frac{6x^6}6+o(x^6)\Bigr)+x^4 \\
&=6x^2-x^4 +\frac{x^6}{20}+o(x^6)-\Bigl(6x^2-x^6+o(x^6)\Bigr)+x^4 \\
&=\frac{x^6}{20}+x^6+o(x^6)=\frac{21x^6}{20}+o(x^6)\sim_{0}\frac{21x^6}{20}.
\end{align}
A: Using the limit $\lim_{x\to 0}(e^x-1)/x=1$ the denominator of the expression under limit can be safely replaced by $x^6$. Next we add and subtract $6x^2$ in numerator and express it as $$6(x^2-\sin x^2)+6x\sin x-6x^2+x^4$$ This allows us to split the desired limit as a sum of two limits the first one of which is $$6\lim_{x\to 0}\frac{x^2-\sin x^2}{x^6}=6\lim_{t\to 0}\frac{t-\sin t} {t^3}=1$$ via L'Hospital's Rule (just applying once) or Taylor series.
The other limit we need to evaluate is $$\lim_{x\to 0}\frac{6\sin x - 6x+x^3}{x^5}$$ We can apply L'Hospital's Rule once to get the expression $$\frac{3x^2-6(1-\cos x)} {5x^4}$$ and putting $x=2t$ this transforms into $$\frac {3}{20}\cdot\frac{t^2-\sin^2t}{t^4}=\frac{3}{20}\cdot\frac{t+\sin t} {t} \cdot\frac{t-\sin t} {t^3}$$ and this tends to $(3/20)(2)(1/6)=1/20$.
The desired limit is thus $1+(1/20)=21/20$. You should see that the limit has been evaluated using just two applications of L'Hospital's Rule.
Also observe that the the term $6x^2$ was added and subtracted in numerator by observing the term $-6\sin x^2$ and knowing that this could lead to a split as $6(x^2-\sin x^2)$. Whenever you see that an expression can be split into multiple terms always try to ensure that at least one of the terms after split has a finite limit. Then that particular term can be handled without knowing anything about limiting behavior of other terms.
