Solveeit Logo
Login

Question

Question: How do you evaluate the limit \( \dfrac{\sin x}{2x} \) as \( x \) approaches 0?...

How do you evaluate the limit sin⁡x2x\dfrac{\sin x}{2x} as xx approaches 0?

Explanation

Solution

Hint : We first try to find the function and approaching value of the variable xx . 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 lim⁡x→a+ f(x)=lim⁡x→a− f(x)=f(a)\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 sin⁡x2x\dfrac{\sin x}{2x} as x→0x\to 0 . Therefore, we need to find lim⁡x→0 sin⁡x2x\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{2x} .
Let’s assume the function as f(x)=sin⁡x2xf\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}^{+}},0,{{0}^{-}} .
0+{{0}^{+}} represents that the value is approaching from the right-side or greater side of the point and 0−{{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
lim⁡x→0+ f(x)=lim⁡x→0− f(x)\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
lim⁡x→a mf(x)=mlim⁡x→a f(x)\underset{x\to a}{\mathop{\lim }}\,mf\left( x \right)=m\underset{x\to a}{\mathop{\lim }}\,f\left( x \right) . Here mm is a constant.
So, lim⁡x→0 sin⁡x2x=12lim⁡x→0 sin⁡xx\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 lim⁡x→0 sin⁡xx=1\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 which gives
lim⁡x→0 sin⁡x2x=12lim⁡x→0 sin⁡xx=12\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 sin⁡x2x\dfrac{\sin x}{2x} as xx approaches 0 is of value 12\dfrac{1}{2} .
So, the correct answer is “ 12\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.