The limit of sin2x/x when x→0 is equal to **2** and the limit of sin2x/x when x→∞ is equal to** 0**. Here we find the limits of sin2x/x when x tends to 0 or ∞.

The formulas of the limits of sin2x/x when x→0 or x→∞ is given below:

- lim
_{x→0}sin2x/x = 2. - lim
_{x→∞}sin2x/x = 0.

## Find lim_{x→0} sin2x/x

**Answer:** The function sin2x/x has the limit **2** when x tends to 0.

**Explanation**:

We have

lim_{x→0} sin2x/x

= lim_{x→0} (sin2x/2x × 2)

= 2 lim_{x→0} sin2x/2x

Let z=2x, so that z→0 when x→0. Then the above limit will be

= 2 lim_{z→0} sinz/z

= 2 × 1 as we know that lim_{x→0} sinx/x = 1.

= 2

So the value of lim x→0 sin2x/x is equal to **2**.

## Find lim_{x→∞} sin2x/x

**Answer:** The function sin2x/x has the limit **0** when x tends to ∞.

**Explanation**:

As sine function takes values in [-1, 1], we have that -1 ≤ sin 2x ≤ 1. As x approaches to ∞, dividing by x it follows that

$-\dfrac{1}{x} \leq \dfrac{\sin 2x}{x} \leq \dfrac{1}{x}$

Now, taking the limit x→∞ we obtain that

$- \lim\limits_{x \to \infty} \dfrac{1}{x}$ $\leq \lim\limits_{x \to \infty} \dfrac{\sin 2x}{x}$ $\leq \lim\limits_{x \to \infty} \dfrac{1}{x}$

⇒ 0 ≤ lim_{x→∞} $\dfrac{\sin 2x}{x}$ ≤ 0

Thus, by the squeeze theorem of limits, the limit of sin2x/x when x approaches to infinity is equal to 0.

**ALSO READ:**

lim_{x→0} (e^{x}-1)/x = 1 | lim_{x→∞} sinx/x = 0 |

lim_{x→0} x/cosx = 0 | lim_{x→0} x/sinx = 1 |

lim_{x→0} tanx/x = 1 | lim_{x→∞} tanx/x = undefined |

lim_{x→0} sin(1/x) = undefined

Epsilon – delta definition of limit

## FAQs

**Q1: What is the limit of sin2x/x as x approaches infinity?**

Answer: The limit of sin2x/x as x approaches to infinity is equal to 0, and this can be proved by the squeeze theorem of limits.

Q2: **What is the limit of sin2x/x as x approaches 0?**

Answer: The value of lim_{x→0} sin2x/x is equal to 2.