Can You Solve These Challenging Calculus Problems?

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SUMMARY

The forum discussion centers on developing and solving challenging calculus problems for a course project, specifically targeting first and second-year calculus topics. Participants suggest focusing on problems involving trigonometric identities and limits, particularly the limit of sin(x)/x as x approaches 0, which equals 1. Resources such as tutorials from Harvey Mudd College are recommended for further exploration of calculus concepts, including Taylor Series and integration techniques. The discussion highlights the importance of understanding limits and the application of trigonometric identities in solving calculus problems.

PREREQUISITES
  • Understanding of first and second-year calculus concepts
  • Familiarity with trigonometric identities and their applications
  • Knowledge of limits, particularly the limit of sin(x)/x
  • Basic integration techniques, including trigonometric substitution
NEXT STEPS
  • Research trigonometric substitution techniques for integrals
  • Explore the Taylor Series and its applications in calculus
  • Study the properties and proofs of limits, especially involving sin(x)/x
  • Practice solving challenging calculus problems from online resources like Harvey Mudd College
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Students in introductory calculus courses, educators seeking problem sets for teaching, and anyone looking to deepen their understanding of calculus concepts and problem-solving techniques.

  • #31
weagle2008 said:
Actually all you need is your trig identities:

sinx / x = cosx ; and since cos (0) = 1 ; sin (0) / 0 = 1

What? Take x = pi.

<br /> \frac{sin \pi}{\pi} = 0<br />

But...

<br /> cos \pi = -1<br />
 
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  • #32
Let t = time when this question was asked.

Find t.

:wink:
 
  • #33
l'Hôpital said:
What? Take x = pi.

<br /> \frac{sin \pi}{\pi} = 0<br />

But...

<br /> cos \pi = -1<br />

The question isn't as x approaches Pi it is as x approaches 0. As anyone knows Pi = 180 degrees. 2*Pi = 360 degrees which is = 0 in trig. Thus cos (Pi) = -1, but cos (0) = 1.
 
  • #34
weagle2008 said:
The question isn't as x approaches Pi it is as x approaches 0. As anyone knows Pi = 180 degrees. 2*Pi = 360 degrees which is = 0 in trig. Thus cos (Pi) = -1, but cos (0) = 1.

But you stated that (sinx)/x = cosx, which means that this is true for all x (which is wrong). This is what he was talking about.
sinx / x = cosx ; and since cos (0) = 1 ; sin (0) / 0 = 1

I would conjecture that you meant \lim_{x\rightarrow 0}\frac{sinx}{x} = \lim_{x\rightarrow 0} cosx. However, I'm not exactly sure how you got to that without L'H.
 

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