Find $\displaystyle \lim_{x \to 0}{{\left( \cos x \right)}^{\dfrac{1}{\sin x}}}=?$
(a) 1
(b) 0
(c) 2
(d) -1

Hint: To solve this problem we are to start with using a different theorem to get our solution. The theorem says if $\displaystyle \lim_{x \to a}f\left( x \right)=\displaystyle \lim_{x \to a}g\left( x \right)=0$ such that $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}$ exists, then $\displaystyle \lim_{x \to a}{{\left[ 1+f\left( x \right) \right]}^{\dfrac{1}{g\left( x \right)}}}={{e}^{\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}}}$ . Then we will consider $f\left( x \right)=\cos x-1$ and $g\left( x \right)=\sin x$, so that the needed form of the theorem can be used. By using trigonometric identities and further simplification we will reach our desired solution.

Complete step by step solution:
According to the problem, to start with, we are to find the value of $\displaystyle \lim_{x \to 0}{{\left( \cos x \right)}^{\dfrac{1}{\sin x}}}$.
So, to get our value we will use the theorem that, if $\displaystyle \lim_{x \to a}f\left( x \right)=\displaystyle \lim_{x \to a}g\left( x \right)=0$ such that $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}$ exists, then $\displaystyle \lim_{x \to a}{{\left[ 1+f\left( x \right) \right]}^{\dfrac{1}{g\left( x \right)}}}={{e}^{\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}}}$
Now, to get our solution, we need to consider, $f\left( x \right)=\cos x-1$ and $g\left( x \right)=\sin x$
And, a = 0.
So, we start with, $\displaystyle \lim_{x \to 0}f\left( x \right)=\displaystyle \lim_{x \to 0}\left( \cos x-1 \right)$
As, cos 0 =1, we get, $\displaystyle \lim_{x \to 0}f\left( x \right)=1-1=0$
Again for the next function, $\displaystyle \lim_{x \to 0}g\left( x \right)=\displaystyle \lim_{x \to 0}\left( \sin x \right)=0$
So, the first condition satisfies.
And we can also easily see, $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}$exists.
So, we can use our given theorem.
From, $\displaystyle \lim_{x \to a}{{\left[ 1+f\left( x \right) \right]}^{\dfrac{1}{g\left( x \right)}}}={{e}^{\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}}}$, putting the value of the f and g function,
$\displaystyle \lim_{x \to a}{\mathop{\Rightarrow \lim }}\,{{\left[ 1+\cos x-1 \right]}^{\dfrac{1}{\sin x}}}={{e}^{\displaystyle \lim_{x \to 0}\left( \dfrac{\cos x-1}{\sin x} \right)}}$
Now, we will try to find the value of $\displaystyle \lim_{x \to 0}\left( \dfrac{\cos x-1}{\sin x} \right)$ .
So, $\displaystyle \lim_{x \to 0}\left( \dfrac{\cos x-1}{\sin x} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{-2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right)$
Using the trigonometric formulas,$\cos 2x-1=-2{{\sin }^{2}}x$ and $\sin 2x=2\sin x\cos x$ .
Simplifying,
$\displaystyle \lim_{x \to 0}{\mathop{\Rightarrow \lim }}\,\left( -\tan \dfrac{x}{2} \right)$
Putting the value of x = 0, we get the value of the limit as 0.
So, now, from ${{e}^{\displaystyle \lim_{x \to 0}\left( \dfrac{\cos x-1}{\sin x} \right)}}$we get,
${{e}^{\displaystyle \lim_{x \to 0}\left( \dfrac{\cos x-1}{\sin x} \right)}}={{e}^{0}}=1$
Then, we get, $\displaystyle \lim_{x \to 0}{{\left( \cos x \right)}^{\dfrac{1}{\sin x}}}=1$

So, the correct answer is “Option A”.

Note: A limit theorem has been used to get the solution of the problem. This is the only way to solve this problem by using this theorem. Otherwise if we try to solve this manually the solution process would be much longer and harder to process. So, to solve this problem we must use this theorem to get the right option.