The limit of sinx/x as x approaches infinity is equal to zero. That is, lim_{x→∞} sinx/x = 0. This can be proved using the Squeeze Theorem of limits which will be done here.

## Limit of sinx/x as x goes to infinity

**Question:** What is the limit of sinx/x when x**→**∞.

**Answer:** The limit of sinx/x when x tends to infinity is equal to 0.

*Proof:*

We know that the value of sinx lies between -1 and 1 for any real number x, that is,

-1 ≤ sinx ≤ 1

Dividing by x, we get that

$-\dfrac{1}{x} \leq \dfrac{\sin x}{x} \leq \dfrac{1}{x}$

Taking limit x**→**∞, we obtain that

$-\lim\limits_{x \to \infty} \dfrac{1}{x} \leq \lim\limits_{x \to \infty} \dfrac{\sin x}{x}$ $\leq \lim\limits_{x \to \infty} \dfrac{1}{x}$

⇒ 0 ≤ $\lim\limits_{x \to \infty} \dfrac{\sin x}{x}$ ≤ 0

As $\lim\limits_{x \to \infty} \dfrac{\sin x}{x}$ lies between 0 and o, it is actually 0. So we obtain that the limit of sinx/x is equal to zero when x goes to infinity. The method used here is known as the Squeeze Theorem of limits.

**Remark:**

- Following the same approach as above, we can prove that the limit of sinx/x
^{2}when x→∞ is also equal to 0. - As -1 ≤ cosx ≤ 1, by the Squeeze Theorem of limits, we get that the limit of cosx/x when x→∞ is equal to 0.

**ALSO READ:**

Limit of x/sinx when x→0 |

Epsilon-delta definition of limits |

Prove that sinx is continuous |

cosx is continuous: proof |

## FAQs

**Q1: What is the limit of sinx/x as x goes to infinity?**

Answer: By Squeeze Theorem, the limit of sinx/x is equal to 0 as x goes to infinity.