# How do you find the limit of (sinx)/(3x) as x approaches oo?

Jun 16, 2016

0

#### Explanation:

you can see sinx as a wave picture whose value is always between 1 and -1
graph{sinx [-10, 10, -5, 5]}

and then see 3x

when x becomes larger and larger in denominator

we dont need to care the value of sinx

it will become 0

Jun 16, 2016

${\lim}_{x \to \pm \infty} \sin \frac{x}{3 x} = 0$

#### Explanation:

$\textcolor{b l u e}{\text{Consider } \sin \left(x\right)}$

This function can assume all values between and including -1 and +1. It will repeat this cycle every $2 \pi$ radians as appropriate for the value of $x$. So $\sin \left(x\right) \in \left(- 1 , + 1\right)$
Where $\left(- 1 , + 1\right)$ is the range of all value between and including -1 to +1.

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$\textcolor{b l u e}{\text{Consider } 3 x}$

As $x$ becomes increasing larger in the positive or negative direction, $\frac{1}{3 x}$ becomes increasingly smaller.

So for $x < 0 \text{ "sin(x)/(3x)" tends to 0 but on the negative side of 0}$

So for $x > 0 \text{ "sin(x)/(3x)" tends to 0 but on the positive side of 0}$

Thus ${\lim}_{x \to \pm \infty} \sin \frac{x}{3 x} = \sin \frac{x}{\infty} = 0$

As you move further and further away from the origin the amplitude lessens and approaches $y = 0$