# Evaluate the limit: lim_{x to 1} {sin(1 - cos x)} / x.

## Question:

Evaluate the limit:

{eq}\displaystyle \lim_{x \to 1} \dfrac {\sin(1 - \cos x)} x {/eq}.

## Limit of a Function:

Let {eq}f {/eq} be a given function. Then a real number {eq}l {/eq} is said to be the limit of the function {eq}f {/eq} at a point {eq}a {/eq} in the domain of {eq}f {/eq} if for each {eq}\epsilon {/eq} there exists a {eq}\delta >0 {/eq} such that {eq}\left| f(x) - l \right| < \epsilon {/eq} for all {eq}x {/eq} satisfying {eq}\left| {x - a} \right| < \delta {/eq}.

Here the given limit is {eq}\displaystyle \mathop {\lim }\limits_{x \to 1} \frac{{\sin (1 - \cos x)}}{x}. {/eq}

Then we have {eq}\displaystyle x \to 1 \Rightarrow \cos \left( x \right) \to \cos \left( 1 \right) \Rightarrow 1 - \cos x \to 1 - \cos \left( 1 \right) \Rightarrow \sin (1 - \cos x) \to \sin \left( {1 - \cos \left( 1 \right)} \right). {/eq}

Using the above, we get

{eq}\displaystyle x \to 1 \Rightarrow \cos \left( x \right) \to \cos \left( 1 \right) \Rightarrow 1 - \cos x \to 1 - \cos \left( 1 \right) \Rightarrow \sin (1 - \cos x) \to \sin \left( {1 - \cos \left( 1 \right)} \right) {/eq}

which is the required value the limit. 