The trigonometric sine function is involved in this function. So, the limit of this function can be evaluated by considering the limit rule of trigonometric functions in this problem.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$ $\,=\,$ $1$

So, let’s try to convert the given function into this form in this limits problem.

$\displaystyle \large \lim_{x \,\to\, \pi}{\normalsize \dfrac{x-\pi}{\sin{x}}}$

In this problem, it is given that $x \to \pi$, then $x-\pi \to 0$ mathematically.

$= \,\,\,$ $\displaystyle \large \lim_{x -\pi \,\to\, 0}{\normalsize \dfrac{x-\pi}{\sin{x}}}$

Take $y = x-\pi$, then $x = \pi+y$. Now, transform the entire limit function in terms of $y$.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y}{\sin{(\pi+y)}}}$

If this function is expressed in reciprocal form then it will be similar to the limit rule of trigonometric function.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{(\pi+y)}}{y}}}$

The limit of a reciprocal function can be written as reciprocal of limit of the function as per reciprocal rule of limits.

$= \,\,\,$ $\dfrac{1}{\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{(\pi+y)}}{y}}}$

The angle in the sine function belongs to third quadrant. In third quadrant, $\sin{(\pi+y)} = -\sin{y}$.

$= \,\,\,$ $\dfrac{1}{\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{-\sin{y}}{y}}}$

$= \,\,\,$ $\dfrac{-1}{\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{y}}{y}}}$

As per limit rule of trigonometric function, the limit of sinx/x as x approaches 0 equals to one. So, the denominator should be equal to one.

$= \,\,\,$ $\dfrac{-1}{1}$

$= \,\,\,$ $-1$

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