How do I find this limit without using L'Hôpital's rule? Finding this limit using  L'Hôpital's rule is easy, but how to do it without using L'Hôpital's rule?
$$\lim_{x \rightarrow 0} \frac{(1+\sin x)^{\csc x}-e^{x+1}}{\sin (3x)}$$
 A: We may proceed as follows and reduce the complicated limit expression to a simple one before applying series expansions
$$\begin{aligned}L&=\lim_{x \to 0}\frac{(1 + \sin x)^{\csc x} - e^{x + 1}}{\sin 3x}\\
&=\lim_{x\to 0}\frac{\exp\{\csc x\log(1+\sin x)\} -\exp(1+x)}{\sin 3x}\\
&=\lim_{x\to 0}\frac{\exp(1+x)\left\{\exp\left(\csc x\log(1+\sin x) -1 -x\right) -1\right\}}{\sin 3x}\\
&=\lim_{x\to 0}\frac{\exp(1+x)\left\{\exp\left(\csc x\log(1+\sin x) -1 -x\right) -1\right\}}{3x}\cdot\frac{3x}{\sin 3x}\\
&=\frac{e}{3}\lim_{x\to 0}\frac{\exp\{\csc x\log(1+\sin x) -1 -x\} -1}{x}\\
&=\frac{e}{3}\lim_{x\to 0}\frac{e^{t} -1}{t}\cdot\frac{t}{x}\\
&=\frac{e}{3}\lim_{x\to 0}\frac{t}{x}\\
&=\frac{e}{3}\lim_{x\to 0}\frac{\csc x\log(1+\sin x)-1-x}{x}\\
&=\frac{e}{3}\left(\lim_{x\to 0}\frac{\log(1+\sin x)-\sin x}{x\sin x}-1\right)\\
&=\frac{e}{3}\left(\lim_{x\to 0}\frac{\log(1+\sin x)-\sin x}{\sin^{2} x}\cdot\frac{\sin x}{x}-1\right)\\
&=\frac{e}{3}\left(\lim_{z\to 0}\frac{\log(1+z)-z}{z^{2}}-1\right)\\
&=\frac{e}{3}\left(\lim_{z\to 0}\dfrac{\left(z - \dfrac{z^{2}}{2} + \cdots\right)-z}{z^{2}}-1\right)\\
&=\frac{e}{3}\cdot\frac{-3}{2}=-\frac{e}{2}\end{aligned}$$
In the above derivation we have $z = \sin x$ and $$\begin{aligned}t&=\csc x\log(1+\sin x) -1-x\\
&= \frac{\log(1 + \sin x)}{\sin x} - 1 - x\\
&= \frac{\log(1 + z)}{z} - 1 - x\end{aligned}$$ so that both $t$ and $z$ tend to $0$ as $x\to 0$.
A: Hint
Taylor expansion built at $x=0$ leads to a solution since $$\csc(x)=\frac{1}{x}+\frac{x}{6}+O\left(x^2\right)$$ $$(1+\sin x)^{\csc x}=e-\frac{e x}{2}+O\left(x^2\right)$$ $$e^{x+1}=e+e x+O\left(x^2\right)$$ $$(1+\sin x)^{\csc x}-e^{x+1}=-\frac{3 e x}{2}+O\left(x^2\right)$$ $$\sin(3x)=3 x+O\left(x^2\right)$$ and then the limit equal to $-\frac{e}{2}$
