The limit of sinx/x as x approaches infinity is denoted by lim_{x→∞} (sinx/x) and its value is 0. So the limit formula of sinx/x as x→∞ is given as follows:

lim_{x→∞} $\dfrac{\sin x}{x}$ = 0.

This limit formula can be proved using the Squeeze theorem of limits.

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## Evaluate lim_{x→∞} (sinx/x)

**Answer:** The value of lim_{x→∞} (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 lim_{x→∞}, we deduce that

lim_{x→∞} $\dfrac{-1}{x}$ ≤ lim_{x→∞} $\dfrac{\sin x}{x}$ ≤ lim_{x→∞} $\dfrac{1}{x}$

⇒ 0 ≤ lim_{x→∞} $\dfrac{\sin x}{x}$ ≤ 0.

So by the Squeeze theorem, it follows that lim_{x→∞} (sinx/x) = 0.

### Therefore, the value of lim_{x→∞} (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/x^{2} when x tends to infinity.

## Limit of sinx/x^{2} as x approaches infinity

Answer: The limit of sinx/x^{2} is 0 when x→∞, that is, lim_{x→∞} (sinx/x^{2}) = 0. |

**Method1:**

lim_{x→∞} (sinx/x^{2}) = lim_{x→∞} $(\dfrac{\sin x}{x} \times \dfrac{1}{x})$

⇒ lim_{x→∞} (sinx/x^{2}) = lim_{x→∞} $\dfrac{\sin x}{x}$ × lim_{x→∞} $\dfrac{1}{x}$ by the properties of limits.

⇒ lim_{x→∞} (sinx/x^{2}) = 0 × 0 = 0, by the above limit formula.

⇒ lim_{x→∞} (sinx/x^{2}) = 0.

So the limit of sinx/x^{2} 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, lim_{x→∞} (sinx/x^{2}) = 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 lim_{x→0} (sinx/x) = 1.

So the limit of xsin(1/x) as x approaches infinity is equal to 1.

## FAQs

**Q1: What is the limit of sinx/x when x tends to infinity?**

Answer: The limit of sinx/x is equal to 0 when x tends to infinity. That is, lim_{x→∞} sinx/x = 0.

**Q2: What is the limit of xsin(1/x) when x tends to infinity?**

Answer: The limit of xsin(1/x) is equal to 1 when x tends to infinity, that is, lim_{x→∞} 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.