Finding $a$ from $\lim\limits_{x\rightarrow0}(1+a\sin x)^{\csc x} =4$ The question is to find the value of $a$ from the following equation:
$$\lim_{x\rightarrow 0}(1+a\sin x)^{\csc x} =4 $$
 A: You have
$$
\lim_{x\rightarrow 0}\left(1+a\sin x\right)^\frac{1}{\sin x}=\\
\lim_{x\rightarrow 0}\left(\left(1+a\sin x\right)^\frac{1}{a\sin x}\right)^a
$$
if you set $t=1/(a\sin x)$, and given that $t$ tends to $\infty$ when $x$ tends to $0$, you have
$$
\lim_{t\rightarrow\infty}\left(\left(1+\frac{1}{t}\right)^{t}\right)^a=e^a\\
$$
A: Hint for your problem: $\lim_{n\rightarrow 0}(1+n)^{1/n}=e$
A: Another possible way to do it: take the logarithm of both sides. Since it's a continuous function, you can put it inside the limit and use the logarithm properties to take $\csc x$ outside and then use L'Hôpital's:
$$\lim_{x \to 0} (1+a \sin x)^{\frac1{\sin x}} = 4$$
$$\lim_{x \to 0} \ln [(1+a \sin x)^{\frac1{\sin x}} ] = \ln 4
$$
$$
\lim_{x \to 0} \frac{\ln(1+a\sin x)}{\sin x} = \ln 4
$$
This last limit is a $0/0$ indetermination, and L'Hôpital's solves it easily.
A: Let A = $ (1+a\sin x)^{\csc x}$
$\log A=\csc x \cdot\log(1+a\sin x)=\dfrac{\log(1+a\sin x)}{\sin x}$
$\lim_{x\rightarrow 0}\log A=a\lim_{x\rightarrow 0}\dfrac{\log(1+a\sin x)}{a\sin x}=a$ as $a\sin x \rightarrow 0\  as\ x\rightarrow 0$
$\implies A=e^a$ which is equal to $4$.
$\implies a=\log_e4$
A: Let $u = a \sin x$. Then
$$
(1+a \sin x)^{\csc x} = (1 + u)^{\frac{a}{u}} = ((1+u)^{\frac{1}{u}})^a
$$
and we note that as $x \to 0$ we have $u \to 0$. Then we want to solve
$$
4 = \lim_{u \to 0} ((1+u)^{\frac{1}{u}})^a = (\lim_{u \to 0}(1+u)^{\frac{1}{u}})^a = e^a.
$$
Taking logarithms of both sides we find $a = \log 4$, where the logarithm is taken base $e$.
