Evaluate the limit, if it exist. lim_{x to 0} {sin 5 x} / {2 x}.

Question:

Evaluate the limit, if it exist.

{eq}\displaystyle \lim_{x \to 0} \dfrac {\sin 5 x} {2 x} {/eq}.

Limits:

We have a function that has a sine term and a linear term. We have many methods to evaluate the limits but here we will use the standard limit formula.

Answer and Explanation:

$$\displaystyle \lim_{x \to 0} \dfrac {\sin 5 x} {2 x}\\ $$

We can write it as:

$$\displaystyle \lim_{x \to 0}\frac{5}{2}\frac{\sin (5x)}{5x}\\ $$

We will use the standard limit formula:

$$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1\\ $$

So we get the answer as:

$$\frac{5}{2} $$

The limit exists and has a finite value:


Learn more about this topic:

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Understanding the Properties of Limits

from Math 104: Calculus

Chapter 6 / Lesson 5
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