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  1. AP Calculus
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Limits and Continuity

Question 1
college-boardCalculus AB/BCAPExam Style
1 mark

A student is trying to solve this problem lim⁡x→0sin⁡5xx\lim_{x \to 0} \frac{\sin 5x}{x}limx→0​xsin5x​. Here is what this student did Step 1: lim⁡x→0sin⁡5xx(55)\lim_{x \to 0} \frac{\sin 5x}{x} \left( \frac{5}{5} \right)limx→0​xsin5x​(55​) Step 2: lim⁡x→05sin⁡5x5x\lim_{x \to 0} \frac{5 \sin 5x}{5x}limx→0​5x5sin5x​ Step 3: 5⋅1=55 \cdot 1 = 55⋅1=5 Is this student correct?

Question 2
college-boardCalculus AB/BCAPExam Style
1 mark

True or False: the limit of a constant a constant.

Question 3
college-boardCalculus AB/BCAPExam Style
1 mark

What type of discontinuity does h(p)=∣p∣−4/(∣p∣−4)h(p)=\sqrt{\left|p\right|-4}/(\left|p\right|-4)h(p)=∣p∣−4​/(∣p∣−4) exhibit as ppp approaches ±4\pm 4±4?

Question 4
college-boardCalculus AB/BCAPExam Style
1 mark

A pair of students are trying to solve this problem: lim⁡x→0sec⁡(2x)\lim_{x \to 0} \sec(2x)limx→0​sec(2x). The following are the steps the students follow: Step 1: lim⁡x→0sec⁡(2x)=lim⁡x→0sec⁡⋅lim⁡x→0(2x)\lim_{x \to 0} \sec(2x) = \lim_{x \to 0} \sec \cdot \lim_{x \to 0} (2x)limx→0​sec(2x)=limx→0​sec⋅limx→0​(2x) Step 2 = 0⋅00 \cdot 00⋅0 Step 3 = 000 Are the students correct?

Question 5
college-boardCalculus AB/BCAPExam Style
1 mark

Find the limit of the function f(x)=3x2−4x+22x2−5x+3f(x) = \frac{3x^2 - 4x + 2}{2x^2 - 5x + 3}f(x)=2x2−5x+33x2−4x+2​ as xxx approaches 1.

Question 6
college-boardCalculus AB/BCAPExam Style
1 mark

Find the lim⁡x→2(5x3−2x2+4x−1)\lim_{x \to 2} (5x^3 - 2x^2 + 4x - 1)limx→2​(5x3−2x2+4x−1).

Question 7
college-boardCalculus AB/BCAPExam Style
1 mark

​​Consider the function f(x)=x+2xf(x) = \sqrt{x} + 2xf(x)=x​+2x. Find the limit as xxx approaches 4.

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Question 8
college-boardCalculus AB/BCAPExam Style
1 mark

What is lim⁡x→−4x2−x−20x+4\lim_{x \to -4} \frac{x^2 - x - 20}{x + 4}limx→−4​x+4x2−x−20​?

Question 9
college-boardCalculus AB/BCAPExam Style
1 mark

Given that lim⁡x→−∞g(x)=L\lim_{{x \to -\infty}} g(x) = Llimx→−∞​g(x)=L and lim⁡h→0g(−1+h)−g(−1)h=m\lim_{{h \to 0}} \frac{g(-1+h)-g(-1)}{h} = mlimh→0​hg(−1+h)−g(−1)​=m, which expression accurately expresses L in terms of m?

Question 10
college-boardCalculus AB/BCAPExam Style
1 mark

When faced with computing lim⁡h→0(a+h)n−anh\lim_{{h \to 0}} \frac{(a+h)^n-a^n}{h}limh→0​h(a+h)n−an​ where nnn is a positive integer, why would expansion using binomial theorem be less favorable than applying another technique?