# Limit of sinx/x as x approaches infinity: Formula, Proof

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.

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

## Evaluate limx→∞ (sinx/x)

Answer: The value of limx→∞ (sinx/x) is 0.

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

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.

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∞.

Put z=1/x. So z0 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

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, limx→∞ 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, limx→∞ xsin(1/x) = 1.

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