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Question: How do you evaluate the limit \(\dfrac{\sin x}{x}\) as \(x \to \pi \)?...

How do you evaluate the limit sin⁡xx\dfrac{\sin x}{x} as x→πx \to \pi ?

Explanation

Solution

As the limit is not zero and the denominator is x so, we will apply the process of substitution here in order to get the correct answer. Also, to solve the question further we will use the formula sin⁡(π)=0\sin \left( \pi \right)=0 in this function. And to simplify the limit we can write it as lim⁡x→π(sin⁡xx)\displaystyle \lim_{x \to \pi }\left( \dfrac{\sin x}{x} \right) in a limit form.

Complete step-by-step answer:
We are going to use the definition of limit in order to solve this question. Since, limit means about the fact that how close the given function is with its given limit. Now, to find the limit of the function sin⁡xx\dfrac{\sin x}{x} as x approaches the point π\pi we need to solve it by substitution.
For this we will consider lim⁡x→π(sin⁡xx)\displaystyle \lim_{x \to \pi }\left( \dfrac{\sin x}{x} \right) and substitute the value of x as π\pi in the function sin⁡xx\dfrac{\sin x}{x}. Therefore we get lim⁡x→π(sin⁡xx)=sin⁡(π)π\displaystyle \lim_{x \to \pi }\left( \dfrac{\sin x}{x} \right)=\dfrac{\sin \left( \pi \right)}{\pi }. As we know that the value of sin⁡(π)=0\sin \left( \pi \right)=0 so, we now have lim⁡x→π(sin⁡xx)=0π ⇒lim⁡x→π(sin⁡xx)=0 \begin{aligned} & \displaystyle \lim_{x \to \pi }\left( \dfrac{\sin x}{x} \right)=\dfrac{0}{\pi } \\\ & \Rightarrow \displaystyle \lim_{x \to \pi }\left( \dfrac{\sin x}{x} \right)=0 \\\ \end{aligned}
Hence, the value of the limit lim⁡x→π(sin⁡xx)=0\displaystyle \lim_{x \to \pi }\left( \dfrac{\sin x}{x} \right)=0 whenever we get x→πx \to \pi .

Note:
This question was an easy one as we only have to substitute the limit directly. Since, the limit is π\pi here, we get the answer correct. But in case we get a limit of 0 the, we cannot put this value directly in the function. This is due to the fact that if we do so, then the denominator of the function sin⁡xx\dfrac{\sin x}{x} will get 0 resulting into the limit of the function as not defined. This is why it is important to check the denominator first that how it is resulting in, and then only we get the idea of the limit of any function. Unless the denominator is not zero, we can put the limit directly in the function and get the required answer.