AP Calculus AB: Solving Limits Question on Continuity

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SUMMARY

The discussion centers on solving the limits question lim_{x→0} \frac{\sqrt{ax + b} – 2}{x} = 1 from an AP Calculus AB test. The correct values for a and b are definitively a = 4 and b = 4, as established through the application of L'Hôpital's rule and the requirement that the numerator approaches zero as x approaches zero. The mentor's disagreement stems from a misunderstanding of the limit's conditions, specifically regarding the behavior of the function as x approaches zero.

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  • Understanding of limits in calculus
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  • Knowledge of continuity in functions
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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: lim_{x→0} \frac{\sqrt{ax + b} – 2}{x} = 1

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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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.
 

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