The limit of sinx/x as x approaches infinity is equal to zero. That is, limx→∞ 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/x2 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.