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Topic: [ap-calculus] A limit problem
Replies: 1   Last Post: Sep 11, 2012 3:30 PM

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Teague, Dan

Posts: 2,314
Registered: 12/6/04
RE: [ap-calculus] A limit problem
Posted: Sep 11, 2012 3:30 PM
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The error in your analysis is in the step lim (x->0) (2x*sin 2x)/(2x * x^3)].

We know that sin(2x)/2x = 1 as x->0, so
lim (x->0) (sin 2x)/(2x )* lim (x->0) 2x/x^3)= lim (x->0) (2x/x^3). You then couple this with the other term lim (x->0) (2/x^2) to get zero.

But, the limit of a product is the product of the limits only if both limits exist, and lim (x->0) (2x/x^3) does not exist.

I find it easier to think of this problem and other limit at zero problems by considering the quadratic approximations. You can do this very early in the calculus class.

sin(2x) behaves like 2x - (2x)^3/6 near x = 0, so (sin 2x - 2x)/x^3 behaves like (2x - 4x^3/3 -2x)/x^3 = 4/3 near x = 0.


Daniel J. Teague
Department of Mathematics
NC School of Science and Mathematics
1219 Broad Street
Durham, NC 27705

-----Original Message-----
From: Roland Cabanos []
Sent: Tuesday, September 11, 2012 10:30 AM
To: AP Calculus
Subject: [ap-calculus] A limit problem

Any help with an explanation to the following problem would be appreciated:

Consider the limit as x approaches 0 for (sin 2x - 2x)/x^3. Graphically, the solution is -4/3. Using L'Hopital's rule also results in the solution of -4/3. However, since I have not taught my students L'Hopital's Rule yet, they are left to other analytic methods. So, lim(x->0) [sin 2x/x^3 - 2x/x^3] = lim (x->0) [(2x*sin 2x)/(2x * x^3) - 2/x^2] = lim(x->0) [2/x^2 - 2/x^2] = lim(x->0) 0 = 0.

Why is the analytic solution different from the graphical and l'Hopital methods?

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