$ \lim_{x\to 0} \left(\frac{5}{2+\sqrt{9+x}}\right)^{\operatorname{cosec} x}$ 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})
&amp;=\ln(2+3\sqrt{1+(x/9)}\,)\\
&amp;=\ln(2+3(1+x/18+o(x)))\\
&amp;=\ln(5+x/6+o(x))\\
&amp;=\ln 5+\ln(1+x/30+o(x))\\
&amp;=\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.
The limit is similar to

$$\lim_{t\to0}(1+t)^{1/t}.$$

In the positives, you can rewrite

$$\lim_{u\to\infty}\left(\frac{u+1}u\right)^u=e.$$

In the negatives,

$$\lim_{u\to-\infty}\left(\frac{u+1}u\right)^u=\lim_{u\to\infty}\left(\frac{u-1}u\right)^{-u}=\lim_{u\to\infty}\left(\frac v{v+1}\right)^{-v-1}
\\=\lim_{v\to-\infty}\left(\frac{v+1}v\right)^v\lim_{v\to\infty}\left(\frac v{v+1}\right)^{-1}=e.$$

So your second reasoning doesn't hold.