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.

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

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

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.