AP Calculus AB: Solving Limits Question on Continuity

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In summary, the conversation involved a tutor and students discussing a test on Limits and Continuity. The students had difficulty with some of the free-response questions, specifically one involving finding values for a and b in a limits problem. The tutor's solution of a = 4 and b = 4 was disputed by their friend/mentor, who stated that x ≠ 0. Ultimately, it was determined that for the function to have a limit, the numerator must approach 0 as x approaches 0, leading to the values of a = 4 and b = 4.
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Chirag B
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I tutor an AP Calculus AB class at the library every week, and I recently gave the students a test on Limits and Continuity. The test was fairly straightforward, but some of the free-response questions were somewhat difficult. The students seemed to get the answers fine, but my friend/mentor seemed to disagree with the answer to one of the questions.

The problem in question is the following limits question: [itex]lim_{x→0}[/itex] [itex]\frac{\sqrt{ax + b} – 2}{x} = 1[/itex]

Given this equality, students were asked to find the values of a and b.

My solutions were a = 4 and b = 4. However, this was disputed by my friend/mentor, who claimed that if x ≠ 0.

Can somebody settle this dispute? What are the values of a and b?
 
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  • #2
In order for the function to have a limit, the numerator must -> 0 as x -> 0. This forces √b = 2 (or b = 4). Using L'Hopital's rule, a = 2√b = 4.
 

FAQ: AP Calculus AB: Solving Limits Question on Continuity

What is a limit in calculus?

A limit in calculus is a mathematical concept that describes the behavior of a function as the input approaches a certain value. It represents the value that the function approaches or "approaches closer and closer" to as the input gets closer to the specified value.

How do you solve a limit question on continuity?

To solve a limit question on continuity, you need to check if the function is continuous at the given point. This means that the limit from both sides of the point should approach the same value. If this is the case, you can simply substitute the value of the point into the function to find the limit. If the function is not continuous, you may need to use other methods such as algebraic manipulation or graphing to determine the limit.

What is the difference between a removable and non-removable discontinuity?

A removable discontinuity, also known as a hole, occurs when there is a point on the graph where the function is undefined, but can be made continuous by filling in the missing point. This can happen when there is a common factor in the numerator and denominator of the function. On the other hand, a non-removable discontinuity occurs when there is a vertical asymptote on the graph, meaning the function cannot be made continuous at that point.

How do you determine if a function is continuous at a point?

To determine if a function is continuous at a point, you need to check three conditions: 1) the limit exists at that point, 2) the function is defined at that point, and 3) the limit and the function value at that point are equal. If all three conditions are met, then the function is continuous at that point.

Can you use direct substitution to solve all limits?

No, direct substitution can only be used to solve limits when the function is continuous at the given point. If the function is not continuous, other methods such as factoring, rationalizing, or using L'Hopital's rule may be necessary to solve the limit.

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