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**Hint:**To determine the limit of $\cot x$ as x approaches ${{\pi }^{-}}$, we are going to first write $\cot x=\dfrac{\cos x}{\sin x}$ and then as x approaches to ${{\pi }^{-}}$ so we are going to take “h” which lies between 0 and 1 and this value of h is very less than 1. Then we write in place of $\left( x-h \right)$ in place of x in $\dfrac{\cos x}{\sin x}$ and then put the limit h tending to 0. And then simplify.

**Complete step-by-step answer:**

In the above problem, it is given that $\cot x$ as x approaches ${{\pi }^{-}}$ so writing this limit expression in the mathematical form we get,

$\Rightarrow \displaystyle \lim_{x\to {{\pi }^{-}}}\cot x$

Now, we know from the trigonometric properties that:

$\cot x=\dfrac{\cos x}{\sin x}$

So, using the above relation in the above limit we get,

$\Rightarrow \displaystyle \lim_{x\to {{\pi }^{-}}}\dfrac{\cos x}{\sin x}$

It is given that x approaches ${{\pi }^{-}}$ so this means that x value is a subtraction of $\pi $ with some number. Let us assume that a small number is “h”. This number “h” lies between 0 and 1 in the following way:

$0 < h << 1$

So, we can write $x=\pi -h$ in the above limit where h is approaching 0.

$\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\cos \left( \pi -h \right)}{\sin \left( \pi -h \right)}$

We know the trigonometric identities of sine and cosine as follows:

$\begin{align}

& \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B; \\

& \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B \\

\end{align}$

Using the above trigonometric identities in the above limit we get,

$\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\cos \pi \cosh +\sin \pi \sinh }{\sin \pi \cosh -\cos \pi \sinh }$

We know the values of sine of $\pi $ and cosine of $\pi $ as follows:

$\begin{align}

& \sin \pi =0; \\

& \cos \pi =-1 \\

\end{align}$

Substituting the above values in the limit expression we get,

$\begin{align}

& \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\left( -1 \right)\cosh +\left( 0 \right)\sinh }{\left( 0 \right)\cosh -\left( -1 \right)\sinh } \\

& \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{-\cosh }{\sinh } \\

\end{align}$

Applying the value of limit by putting h as 0 in the above fraction we get,

$-\dfrac{\cos 0}{\sin 0}$

We know that the value of $\sin 0=0\And \cos 0=1$ so substituting these values in the above w eget,

$-\dfrac{1}{0}$

And we know that $\dfrac{1}{0}$ is not defined or infinity.

$-\infty $

Hence, the evaluation of the above limit is $-\infty $.

**Note:**The possible mistake in the above problem is that in the last two steps you might forget to put a negative sign in front of the infinity because you might think what difference does it make if we remove this negative sign in front of the infinity.

$\begin{align}

& -\dfrac{1}{0} \\

& =\infty \\

\end{align}$

This is the wrong answer because positive and negative infinities are completely different. As one is pointing in a positive direction and the other is in a negative direction so make sure you won’t make this mistake.

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