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  1. AP Calculus
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Explain the first step when evaluating limits using L'Hôpital's Rule.

First, verify that the limit is in an indeterminate form, either 00\frac{0}{0}00​ or ∞∞\frac{\infty}{\infty}∞∞​.

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Explain the first step when evaluating limits using L'Hôpital's Rule.

First, verify that the limit is in an indeterminate form, either 00\frac{0}{0}00​ or ∞∞\frac{\infty}{\infty}∞∞​.

What must you do before applying L'Hôpital's Rule in a Free Response Question?

Show that the limit of the numerator and the limit of the denominator both approach 0 or both approach ±∞\pm \infty±∞.

What does it mean to 'verify conditions' for L'Hopital's Rule?

It means showing that both lim⁡x→af(x)=0\lim_{x \to a} f(x) = 0limx→a​f(x)=0 and lim⁡x→ag(x)=0\lim_{x \to a} g(x) = 0limx→a​g(x)=0 or that both limits approach ±∞\pm \infty±∞.

When can you apply L'Hôpital's Rule multiple times?

If after applying L'Hôpital's Rule once, the limit is still in an indeterminate form, you can apply it again.

How to evaluate lim⁡x→af(x)g(x)\lim_{x\to a}\frac{f(x)}{g(x)}limx→a​g(x)f(x)​ using L'Hôpital's Rule?

  1. Check if the limit is in indeterminate form. 2. Verify conditions: lim⁡x→af(x)=0\lim_{x \to a} f(x) = 0limx→a​f(x)=0 and lim⁡x→ag(x)=0\lim_{x \to a} g(x) = 0limx→a​g(x)=0 or both approach ±∞\pm \infty±∞. 3. Apply L'Hôpital's Rule: lim⁡x→af′(x)g′(x)\lim_{x\to a}\frac{f'(x)}{g'(x)}limx→a​g′(x)f′(x)​. 4. Evaluate the new limit.

Steps to solve lim⁡x→0sin⁡(3x)x\lim_{x \to 0} \frac{\sin(3x)}{x}limx→0​xsin(3x)​ using L'Hopital's Rule.

  1. Check indeterminate form: 00\frac{0}{0}00​. 2. Apply L'Hopital's Rule: lim⁡x→03cos⁡(3x)1\lim_{x \to 0} \frac{3\cos(3x)}{1}limx→0​13cos(3x)​. 3. Evaluate: 3cos⁡(0)=33\cos(0) = 33cos(0)=3.

How to evaluate lim⁡x→∞3x2−87x2+21\lim_{x\to \infty} \frac{3x^2 - 8}{7x^2 + 21}limx→∞​7x2+213x2−8​?

  1. Check indeterminate form: ∞∞\frac{\infty}{\infty}∞∞​. 2. Apply L'Hôpital's Rule: lim⁡x→∞6x14x\lim_{x\to \infty} \frac{6x}{14x}limx→∞​14x6x​. 3. Simplify: 614=37\frac{6}{14} = \frac{3}{7}146​=73​.

What is L'Hôpital's Rule?

If lim⁡x→af(x)g(x)\lim_{x\to a}\frac{f(x)}{g(x)}limx→a​g(x)f(x)​ is of the form 00\frac{0}{0}00​ or ∞∞\frac{\infty}{\infty}∞∞​, then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}limx→a​g(x)f(x)​=limx→a​g′(x)f′(x)​.