Evaluate the limit, if it exist. lim_{x to 0} {sin (pi - x)} / {square root {x^2 - x + 1}}.

Question:

Evaluate the limit, if it exist.

{eq}\displaystyle \lim_{x \to 0} \dfrac {\sin (\pi - x)} {\sqrt {x^2 - x + 1}} {/eq}.

The Limit in Calculus:

If the limit of a function {eq}f(x) {/eq} can be writen as: {eq}\displaystyle\lim_{x\rightarrow \, a} f(x) {/eq}. The limit of a function is used to analyze the behavior of a function near the input point.

To solve this problem, we'll apply direct substitution, which is process of directly substitute the given value of {eq}x {/eq} and use the trigonometric value {eq}\sin (\pi )= 0 {/eq} and simplify the answer.

Answer and Explanation:

We are given:

{eq}\displaystyle \lim_{x \to 0} \dfrac {\sin (\pi - x)} {\sqrt {x^2 - x + 1}} {/eq}


Plug in the value of {eq}x=0 {/eq}

{eq}=\displaystyle \lim_{x \to 0} \dfrac {\sin (\pi - 0)} {\sqrt {0^2 - 0 + 1}} {/eq}

{eq}=\displaystyle \dfrac {\sin (\pi )} {\sqrt {1}} {/eq}


Using the trigonometric value {eq}\sin (\pi )= 0 {/eq}

{eq}=\displaystyle \dfrac {0} {1} {/eq}

{eq}=\displaystyle 0 {/eq}


Therefore, the solution is:

{eq}\displaystyle {\boxed{ \lim_{x \to 0} \dfrac {\sin (\pi - x)} {\sqrt {x^2 - x + 1}} = 0}} {/eq}


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Understanding the Properties of Limits

from Math 104: Calculus

Chapter 6 / Lesson 5
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