The limit of sinx/x as x approaches infinity is denoted by limx→∞ (sinx/x) and its value is 0. So the limit formula of sinx/x as x→∞ is given as follows:
limx→∞ $\dfrac{\sin x}{x}$ = 0.
![Limit of sinx/x as x approaches infinity](https://www.mathstoon.com/wp-content/uploads/2024/07/Limit-of-sinxx-as-x-tends-to-infinity.webp)
This limit formula can be proved using the Squeeze theorem of limits (also known as the Sandwich Theorem).
Table of Contents
Evaluate limx→∞ (sinx/x)
Answer: The value of limx→∞ (sinx/x) is 0.
Explanation:
For all real values of x, we know that
-1 ≤ sin x ≤ 1.
So for large values x, we deduce that
$-\dfrac{1}{x} \leq \dfrac{\sin x}{x} \leq \dfrac{1}{x}$.
Now taking limx→∞, we deduce that
limx→∞ $\dfrac{-1}{x}$ ≤ limx→∞ $\dfrac{\sin x}{x}$ ≤ limx→∞ $\dfrac{1}{x}$
⇒ 0 ≤ limx→∞ $\dfrac{\sin x}{x}$ ≤ 0.
So by the Squeeze theorem, it follows that limx→∞ (sinx/x) = 0.
Therefore, the value of limx→∞ (sinx/x) is equal to 0, proved by the Squeeze/Sandwich theorem of limits.
Now, we will use this limit formula to find the limit of sinx/x2 when x tends to infinity.
Limit of sinx/x2 as x approaches infinity
Answer: The limit of sinx/x2 is 0 when x→∞, that is, limx→∞ (sinx/x2) = 0. |
Method1:
limx→∞ (sinx/x2) = limx→∞ $(\dfrac{\sin x}{x} \times \dfrac{1}{x})$
⇒ limx→∞ (sinx/x2) = limx→∞ $\dfrac{\sin x}{x}$ × limx→∞ $\dfrac{1}{x}$ by the properties of limits.
⇒ limx→∞ (sinx/x2) = 0 × 0 = 0, by the above limit formula.
⇒ limx→∞ (sinx/x2) = 0.
So the limit of sinx/x2 as x approaches infinity is equal to 0.
Method2:
As -1 ≤ sin x ≤ 1, for large values of x we obtain that
$-\dfrac{1}{x^2} \leq \dfrac{\sin x}{x^2} \leq \dfrac{1}{x^2}$.
Taking limit x tends to ∞, we deduce that
$0 \leq \lim\limits_{x \to \infty} \dfrac{\sin x}{x^2} \leq 0$.
So by Sandwich theorem, limx→∞ (sinx/x2) = 0.
Limit of sin2x/x as x approaches infinity
Put z=2x. So z→∞ as x→∞.
Then, $\lim\limits_{x \to \infty} \dfrac{\sin 2x}{x}$
= $\lim\limits_{z \to \infty} \dfrac{\sin z}{z/2}$
= 2 $\lim\limits_{z \to \infty} \dfrac{\sin z}{z}$
= 2 × 0 = 0, by the above limit formula.
Hence the limit of sin2x/x as x approaches infinity is equal to 0.
Related Topics:
Limit of x sin1/x as x approaches infinity
Question: Find the limit of $x \sin \frac{1}{x}$ as x→∞.
Answer:
Put z=1/x. So z→0 when x→∞.
Now, $\lim\limits_{x \to \infty} x \sin \frac{1}{x}$
= $\lim\limits_{x \to \infty} \dfrac{\sin \frac{1}{x}}{\frac{1}{x}}$
= $\lim\limits_{z \to 0} \dfrac{\sin z}{z}$
= 1, by the limit formula limx→0 (sinx/x) = 1.
So the limit of xsin(1/x) as x approaches infinity is equal to 1.
FAQs
Answer: The limit of sinx/x is equal to 0 when x tends to infinity. That is, limx→∞ sinx/x = 0.
Answer: The limit of xsin(1/x) is equal to 1 when x tends to infinity, that is, limx→∞ xsin(1/x) = 1.
This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.