Indeterminate Forms

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Step 1: Simplify the fraction to x. Step 2: Take the limit as x→0. lim_{x→0} x = 0. Answer: 0 Why this matters: This is a 0/0 form only in appearance; after simplification, the limit is easy to find.

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Step 1: Cancel x for x ≠ 0. (x)/(x) = 1. Step 2: Take the limit. lim_{x→0} 1 = 1. Answer: 1 This shows that a 0/0 form can give different answers depending on the functions involved.

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Step 1: Rewrite the fraction as 1/x. Step 2: Observe the behavior as x→0. 1/x grows without bound in magnitude. Answer: ∞ This is another 0/0 form, but the limit is different from the previous example…

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Use it when the limit gives 0/0 or ∞/∞ form. Method: 1. Differentiate the numerator. 2. Differentiate the denominator. 3. Take the new limit. Example: lim_{x→0} [sin(ax)/(bx)] Differentiate both parts…

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Step 1: At x=1, both numerator and denominator become 0, so this is 0/0 form. Step 2: Apply L'Hôpital's Rule. Derivative of numerator = 15x^14. Derivative of denominator = 10x^9. Step 3: Evaluate at x…

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Step 1: At x=0, numerator = 0 and denominator = 0, so this is 0/0 form. Step 2: Apply L'Hôpital's Rule once. Derivative of numerator = sin x. Derivative of denominator = 2x. New limit: lim_{x→0} [sin …

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Step 1: Recognize 0/0 form. Step 2: Apply L'Hôpital's Rule. Derivative of sin 11x = 11 cos 11x. Derivative of 7x = 7. Step 3: Substitute x=0. lim_{x→0} [11 cos 11x / 7] = 11/7. Answer: 11/7…

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Step 1: At x=0, both parts give 0, so use L'Hôpital's Rule. First differentiation: Derivative of cos 2x - 1 = -2 sin 2x. Derivative of cos x - 1 = -sin x. New limit: lim_{x→0} [2 sin 2x/sin x]. Step 2…