AP Calc AB Review for QZ 21-24

• 1

• 5

• Does not exist

Answer explanation

Using the limit property, \( \lim_{x \to 0} \frac{\sin kx}{x} = k \). Here, \( k = 5 \), so \( \lim_{x \to 0} \frac{\sin 5x}{x} = 5 \). Thus, the correct answer is 5.

Tags

l'Hopitâl's Rule

• 4

• 0

Answer explanation

To evaluate \(\lim_{x\rightarrow4}\ \frac{\sqrt{x}-2}{x-4}\), we can use L'Hôpital's rule since it results in the indeterminate form \(\frac{0}{0}\). Differentiating the numerator and denominator gives \(\frac{1}{2\sqrt{x}}\) and 1, respectively. Evaluating at \(x=4\) yields \(\frac{1}{4}\).

Tags

l'Hopitâl's Rule

• 2

• -2

Answer explanation

To find the limit, simplify the expression: \( \frac{\frac{x}{x+2}-2}{x+4} = \frac{\frac{x-2(x+2)}{x+2}}{x+4} = \frac{\frac{-x-4}{x+2}}{x+4} \). As \( x \to -4 \), this approaches \( \frac{1}{2} \). Thus, the answer is \( \frac{1}{2} \).

Tags

l'Hopitâl's Rule

• Does not exist

• 1

• -1

• 0

Answer explanation

To find \( \lim_{x\rightarrow0}\ \frac{\sin(\sin x)}{\sin x} \), we use L'Hôpital's rule. As \( x \to 0 \), both the numerator and denominator approach 0. Differentiating gives \( \frac{\cos(\sin x) \cos x}{\cos x} \), which simplifies to \( \cos(\sin x) \). Evaluating at 0 yields 1.

Tags

l'Hopitâl's Rule

• 1

• 0

• Nonexistent

Answer explanation

Using L'Hôpital's Rule, we differentiate the numerator and denominator. The limit simplifies to \(\frac{6}{5}\) as \(x\) approaches 0, confirming the correct answer is \(\frac{6}{5}\).

Tags

l'Hopitâl's Rule

• 2

• 1

• infinity

• 0

• undefined

Answer explanation

To evaluate \(\lim_{x\rightarrow0}\ \frac{\sin x\cos x-\sin x}{x^2}\), we simplify the expression to \(\frac{\sin x(\cos x-1)}{x^2}\). As \(x\rightarrow0\), both \(\sin x\) and \(\cos x-1\) approach 0, leading to the limit being 0.

Tags

l'Hopitâl's Rule

• 0

• 1

• 2

• DNE

Answer explanation

To find the limit as x approaches 1, substitute x=1 into the expression. The numerator becomes 0 and the denominator also becomes 0, indicating a 0/0 form. Factoring gives (3x^2)(x-1)/(x-1)(x^2-2). Canceling (x-1) leads to 0, confirming the limit is 0.

Tags

l'Hopitâl's Rule

• 0

• 1

• 2

• DNE

Answer explanation

To find the limit, substitute x=0: \lim_{x\rightarrow0} \frac{3x + \sin x}{2x} = \frac{3(0) + \sin(0)}{2(0)}. Using L'Hôpital's Rule, we differentiate the numerator and denominator, yielding \lim_{x\rightarrow0} \frac{3 + \cos x}{2} = \frac{3 + 1}{2} = 2.

Tags

l'Hopitâl's Rule

• 3

• 1

Answer explanation

Using the limit property, \( \lim_{x \to 0} \frac{\sin kx}{kx} = 1 \), we can rewrite the limit as \( \frac{6}{2} = 3 \). Thus, the limit evaluates to 3.

Tags

l'Hopitâl's Rule

• 1

• -1

Answer explanation

To evaluate \( \lim_{x\rightarrow1}\left(\frac{\tan\left(x-1\right)}{x^3-1}\right) \), apply L'Hôpital's Rule since both the numerator and denominator approach 0. Differentiating gives \( \frac{\sec^2(0)}{3x^2} \) evaluated at \( x=1 \), resulting in \( \frac{1}{3} \).

Tags

l'Hopitâl's Rule