Question

How do you evaluate the limit $ \dfrac{\sin x}{2x} $ as $ x $ approaches 0?

Answer 1

Hint: We first try to find the function and approaching value of the variable $ x $ . Then we find the definition of limit and how it applies for the function to find the limit value. The limit only exists when the left-hand and right-hand each limit gives equal value. The mathematical form being $ \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) $ .

Complete step-by-step answer:
We need to find the limit of $ \dfrac{\sin x}{2x} $ as $ x\to 0 $ . Therefore, we need to find $ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{2x} $ .
Let’s assume the function as $ f\left( x \right)=\dfrac{\sin x}{2x} $ .
For our given limit the value of variable x tends to the point 0. This means the value can be approaching from the both sides of the point of 0. We can break it into three parts of $ {{0}^{+}},0,{{0}^{-}} $ .
 $ {{0}^{+}} $ represents that the value is approaching from the right-side or greater side of the point and $ {{0}^{-}} $ represents that the value is approaching from the left-side or lesser side of the point. There is also the fixed point of 2.
Now the limit value will exist only when
 $ \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right) $ .
We need to find the values of the given function for the approaching value of x.
We know that
 $ \underset{x\to a}{\mathop{\lim }}\,mf\left( x \right)=m\underset{x\to a}{\mathop{\lim }}\,f\left( x \right) $ . Here $ m $ is a constant.
So, $ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{2x}=\dfrac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x} $ .
We also know the identity that $ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 $ which gives
 $ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{2x}=\dfrac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=\dfrac{1}{2} $
Therefore, the limit $ \dfrac{\sin x}{2x} $ as $ x $ approaches 0 is of value $ \dfrac{1}{2} $ .
So, the correct answer is “ $ \dfrac{1}{2} $ ”.

Note: The precise definition of a limit is something we use as a proof for the existence of a limit. When we’re evaluating a limit, we’re looking at the function as it approaches a specific point. we approach a particular value of x, the function itself gets closer and closer to a particular value.