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**Hint:**To find the limit of the given expression, we are going to check first of all, if the given expression is in the $\dfrac{0}{0}or\dfrac{\infty }{\infty }$ form. If the given expression is not in this form then we are going to just put the value of the limit of x in the given expression and then solve it. In this way, we can evaluate the limit of the expression.

**Complete step-by-step solution:**

The expression given in the above problem is as follows:

$\displaystyle \lim_{x \to \pi }\left( x-\dfrac{22}{7} \right)$

Now, first of all, we are going to put this value of limit of x in the above expression and see if we are getting the $\dfrac{0}{0}or\dfrac{\infty }{\infty }$ form.

Substituting the value of x as $\pi $ in the above expression we get,

$\pi -\dfrac{22}{7}$

We know that the value of $\pi =\dfrac{22}{7}$ so substituting this value of $\pi $ in the above expression we get,

$\dfrac{22}{7}-\dfrac{22}{7}$

As we are getting the same number with a negative sign then $\dfrac{22}{7}$ will get canceled out and we get,

$=0$

Now, as you can see that we are not getting any $\dfrac{0}{0}or\dfrac{\infty }{\infty }$ form so the final answer of the limit is calculated by putting the given value of the limit of x in the given expression.

In the above solution, we have already put the value of the limit of x in the given expression so the value of the limit of the given expression is 0.

**Hence, we have evaluated the limit as 0.**

**Note:**In this problem, the limit evaluation is pretty straightforward but if you get expressions in the following way then the way to solve them is as follows:

$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}$

When we put the value of x as 0 in the given expression then the evaluation of limit which we got is as follows:

$\begin{align}

& \dfrac{\sin 0}{0} \\

& =\dfrac{0}{0} \\

\end{align}$

We got the $\dfrac{0}{0}$ form and to find the limit of this expression we are going to differentiate the numerator and the denominator of the above expression with respect to x.

We know that the differentiation of $\sin x$ with respect to x is:

$\dfrac{d\sin x}{dx}=\cos x$

And also, the differentiation of x with respect to x is 1.

$\dfrac{dx}{dx}=1$

Now, writing the differentiation of the numerator and the denominator of $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}$ we get,

$\dfrac{\cos x}{1}$

Putting $x=0$ in the above expression we get,

$\begin{align}

& \cos 0 \\

& =1 \\

\end{align}$

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