(3) lim_(x rarr0)(sqrt(2-cos x)-1)/(3x^(2))=

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(3)  lim_(x rarr0)(sqrt(2-cos x)-1)/(3x^(2))=

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Recognize Indeterminate Form: To solve this limit, we first need to recognize that direct substitution of x = 0 into the expression would result in the indeterminate form 0/0. Therefore, we need to apply a technique to simplify the expression or use L'Hôpital's Rule. In this case, we can try to simplify the expression first by multiplying the numerator and denominator by the conjugate of the numerator to eliminate the square root.Multiply by Conjugate: Multiply the numerator and denominator by the conjugate of the numerator, which is (sqrt(2-cos x)+1), to rationalize the numerator. $$ \lim_{x \to 0} \frac{\sqrt{2-cos x}-1}{3x^2} \cdot \frac{\sqrt{2-cos x}+1}{\sqrt{2-cos x}+1} $$Simplify Numerator: After multiplying, we get: $$ \lim_{x \to 0} \frac{(2-cos x)-1^2}{3x^2(\sqrt{2-cos x}+1)} $$ Simplify the numerator: $$ \lim_{x \to 0} \frac{2-cos x-1}{3x^2(\sqrt{2-cos x}+1)} $$ $$ \lim_{x \to 0} \frac{1-cos x}{3x^2(\sqrt{2-cos x}+1)} $$Apply Trigonometric Identity: Now, we can apply the trigonometric identity $cos^2(x) + sin^2(x) = 1$ to recognize that $1 - cos x = sin^2(x)$. This gives us: $$ \lim_{x \to 0} \frac{sin^2(x)}{3x^2(\sqrt{2-cos x}+1)} $$Cancel x^2: We can now simplify the expression by canceling out $x^2$ in the numerator and denominator: $$ \lim_{x \to 0} \frac{sin^2(x)/x^2}{3(\sqrt{2-cos x}+1)} $$Use Trigonometric Limit: We know that $\lim_{x \to 0} \frac{sin(x)}{x} = 1$, so we can use this to further simplify the expression: $$ \lim_{x \to 0} \frac{1}{3(\sqrt{2-cos x}+1)} $$ Since $x$ is approaching 0, $cos x$ approaches $cos(0) = 1$, and the square root in the denominator approaches $\sqrt{2-1} = \sqrt{1} = 1$.Substitute Limit: Substitute the limit into the expression: $$ \frac{1}{3(\sqrt{2-1}+1)} = \frac{1}{3(1+1)} = \frac{1}{3(2)} = \frac{1}{6} $$

$\lim_{x \to 0}\frac{\sqrt{2-\cos x}-1}{3x^{2}}=\square$

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$\lim_{x \to 0}\frac{\sqrt{2-\cos x}-1}{3x^{2}}=\square$