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:
- limx→0 sin2x/x = 2.
- limx→∞ sin2x/x = 0.
Find limx→0 sin2x/x
Answer: The function sin2x/x has the limit 2 when x tends to 0.
Explanation:
We have
limx→0 sin2x/x
= limx→0 (sin2x/2x × 2)
= 2 limx→0 sin2x/2x
Let z=2x, so that z→0 when x→0. Then the above limit will be
= 2 limz→0 sinz/z
= 2 × 1 as we know that limx→0 sinx/x = 1.
= 2
So the value of lim x→0 sin2x/x is equal to 2.
Find limx→∞ 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 ≤ limx→∞ $\dfrac{\sin 2x}{x}$ ≤ 0
Thus, by the squeeze theorem of limits, the limit of sin2x/x when x approaches to infinity is equal to 0.
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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 limx→0 sin2x/x is equal to 2.