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Question: $\lim_{x \to 0} \frac{(1+sinx)^{\frac{1}{2024}} - (1 - sinx)^{\frac{1}{2024}}}{x}$...

lim⁡x→0(1+sinx)12024−(1−sinx)12024x\lim_{x \to 0} \frac{(1+sinx)^{\frac{1}{2024}} - (1 - sinx)^{\frac{1}{2024}}}{x}

Answer

1/1012

Explanation

Solution

The given limit is of the form 00\frac{0}{0} as x→0x \to 0. We can solve this using L'Hopital's Rule, standard limit formulas, or binomial approximation.

Method 1: Using L'Hopital's Rule Let f(x)=(1+sin⁡x)12024−(1−sin⁡x)12024f(x) = (1+\sin x)^{\frac{1}{2024}} - (1 - \sin x)^{\frac{1}{2024}} and g(x)=xg(x) = x. Then f′(x)=ddx[(1+sin⁡x)12024−(1−sin⁡x)12024]f'(x) = \frac{d}{dx} \left[ (1+\sin x)^{\frac{1}{2024}} - (1 - \sin x)^{\frac{1}{2024}} \right] Using the chain rule, dduun=nun−1dudx\frac{d}{du} u^n = n u^{n-1} \frac{du}{dx}: f′(x)=12024(1+sin⁡x)12024−1⋅cos⁡x−12024(1−sin⁡x)12024−1⋅(−cos⁡x)f'(x) = \frac{1}{2024}(1+\sin x)^{\frac{1}{2024}-1} \cdot \cos x - \frac{1}{2024}(1-\sin x)^{\frac{1}{2024}-1} \cdot (-\cos x) f′(x)=12024(1+sin⁡x)−20232024cos⁡x+12024(1−sin⁡x)−20232024cos⁡xf'(x) = \frac{1}{2024}(1+\sin x)^{-\frac{2023}{2024}} \cos x + \frac{1}{2024}(1-\sin x)^{-\frac{2023}{2024}} \cos x g′(x)=ddx(x)=1g'(x) = \frac{d}{dx}(x) = 1

Now, apply L'Hopital's Rule: lim⁡x→0f′(x)g′(x)=lim⁡x→0[12024(1+sin⁡x)−20232024cos⁡x+12024(1−sin⁡x)−20232024cos⁡x]\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \left[ \frac{1}{2024}(1+\sin x)^{-\frac{2023}{2024}} \cos x + \frac{1}{2024}(1-\sin x)^{-\frac{2023}{2024}} \cos x \right] Substitute x=0x=0: =12024(1+sin⁡0)−20232024cos⁡0+12024(1−sin⁡0)−20232024cos⁡0= \frac{1}{2024}(1+\sin 0)^{-\frac{2023}{2024}} \cos 0 + \frac{1}{2024}(1-\sin 0)^{-\frac{2023}{2024}} \cos 0 =12024(1+0)−20232024⋅1+12024(1−0)−20232024⋅1= \frac{1}{2024}(1+0)^{-\frac{2023}{2024}} \cdot 1 + \frac{1}{2024}(1-0)^{-\frac{2023}{2024}} \cdot 1 =12024(1)+12024(1)= \frac{1}{2024}(1) + \frac{1}{2024}(1) =22024=11012= \frac{2}{2024} = \frac{1}{1012}

Method 2: Using Standard Limit Formula and Binomial Approximation We know the standard limit formula: lim⁡y→0(1+y)n−1y=n\lim_{y \to 0} \frac{(1+y)^n - 1}{y} = n. Let n=12024n = \frac{1}{2024}. The expression can be rewritten as: lim⁡x→0[(1+sin⁡x)n−1]−[(1−sin⁡x)n−1]x\lim_{x \to 0} \frac{[(1+\sin x)^n - 1] - [(1 - \sin x)^n - 1]}{x} =lim⁡x→0(1+sin⁡x)n−1x−lim⁡x→0(1−sin⁡x)n−1x= \lim_{x \to 0} \frac{(1+\sin x)^n - 1}{x} - \lim_{x \to 0} \frac{(1 - \sin x)^n - 1}{x}

For the first term: lim⁡x→0(1+sin⁡x)n−1x=lim⁡x→0(1+sin⁡x)n−1sin⁡x⋅sin⁡xx\lim_{x \to 0} \frac{(1+\sin x)^n - 1}{x} = \lim_{x \to 0} \frac{(1+\sin x)^n - 1}{\sin x} \cdot \frac{\sin x}{x} As x→0x \to 0, sin⁡x→0\sin x \to 0. Let y=sin⁡xy = \sin x. =lim⁡y→0(1+y)n−1y⋅lim⁡x→0sin⁡xx= \lim_{y \to 0} \frac{(1+y)^n - 1}{y} \cdot \lim_{x \to 0} \frac{\sin x}{x} =n⋅1=n= n \cdot 1 = n

For the second term: lim⁡x→0(1−sin⁡x)n−1x=lim⁡x→0(1+(−sin⁡x))n−1−sin⁡x⋅−sin⁡xx\lim_{x \to 0} \frac{(1 - \sin x)^n - 1}{x} = \lim_{x \to 0} \frac{(1 + (-\sin x))^n - 1}{-\sin x} \cdot \frac{-\sin x}{x} As x→0x \to 0, −sin⁡x→0-\sin x \to 0. Let z=−sin⁡xz = -\sin x. =lim⁡z→0(1+z)n−1z⋅lim⁡x→0−sin⁡xx= \lim_{z \to 0} \frac{(1+z)^n - 1}{z} \cdot \lim_{x \to 0} \frac{-\sin x}{x} =n⋅(−1)=−n= n \cdot (-1) = -n

Combining the two terms: The limit is n−(−n)=2nn - (-n) = 2n. Substitute n=12024n = \frac{1}{2024}: 2⋅12024=22024=110122 \cdot \frac{1}{2024} = \frac{2}{2024} = \frac{1}{1012}

Alternatively, using binomial approximation (1+u)n≈1+nu(1+u)^n \approx 1+nu for small uu: As x→0x \to 0, sin⁡x→0\sin x \to 0. (1+sin⁡x)12024≈1+12024sin⁡x(1+\sin x)^{\frac{1}{2024}} \approx 1 + \frac{1}{2024} \sin x (1−sin⁡x)12024≈1+12024(−sin⁡x)=1−12024sin⁡x(1-\sin x)^{\frac{1}{2024}} \approx 1 + \frac{1}{2024} (-\sin x) = 1 - \frac{1}{2024} \sin x

Substitute these into the expression: lim⁡x→0(1+12024sin⁡x)−(1−12024sin⁡x)x\lim_{x \to 0} \frac{(1 + \frac{1}{2024} \sin x) - (1 - \frac{1}{2024} \sin x)}{x} =lim⁡x→01+12024sin⁡x−1+12024sin⁡xx= \lim_{x \to 0} \frac{1 + \frac{1}{2024} \sin x - 1 + \frac{1}{2024} \sin x}{x} =lim⁡x→02⋅12024sin⁡xx= \lim_{x \to 0} \frac{2 \cdot \frac{1}{2024} \sin x}{x} =lim⁡x→011012sin⁡xx= \lim_{x \to 0} \frac{1}{1012} \frac{\sin x}{x} Since lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1: =11012⋅1=11012= \frac{1}{1012} \cdot 1 = \frac{1}{1012}

All methods yield the same result.