I solved this question in two different ways and got two different answers to it. Method #1: The question is of the form $1^\infty$. So, to solve it, I took it in the power of e and wrote the numerator in the natural logarithmic function.
Now, according to the Maclauren series, $$\lim_{x\to 0} \left(\frac{\ln (x+1)}{x}\right) = 1$$ Using this result, then rationalising the power and then using the result $$\lim_{x\to 0} \left(\frac{\sin x}{x}\right) = 1$$ we get the answer as $e^{-1/30}$
Method #2:
In this method, I decided to go by the basics instead. I took out the left-hand limit and the right-hand limit separately.
If we want the right-hand limit, then we will have to put $x = 0^+$. Upon doing so, the denominator becomes slightly greater than the numerator, and we can say that the part inside the parenthesis is slightly less than 1. Now $cosec 0^+$ tends to $\infty$ Also, a number less than 1 raised to the power of infinity tends to 0.
$\therefore$ The right-hand limit is zero.
In a similar manner when we take out the left-hand limit, that also comes out to be zero.
Hence, by this method, the answer should be 0.
Please help.
You need to work out what happens to \begin{align} \ln(2+\sqrt{9+x}) &=\ln(2+3\sqrt{1+(x/9)}\,)\\ &=\ln(2+3(1+x/18+o(x)))\\ &=\ln(5+x/6+o(x))\\ &=\ln 5+\ln(1+x/30+o(x))\\ &=\ln 5+x/30+o(x) \end{align} Thus the logarithm of your function is $$ \frac{\ln 5-\ln(2+\sqrt{9+x})}{\sin x}= \frac{\ln 5-\ln 5-x/30+o(x)}{x+o(x)} $$ and the limit (of the logarithm) is $-1/30$. So the given limit is $e^{-1/30}$. Good job with method #1.
Your method #2 is flawed: for instance $$ \lim_{x\to0^+}(1-x)^{1/x}=e^{-1} $$ and not $0$ as you claim.