- #31
l'Hôpital
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weagle2008 said:
What? Take x = pi.
<br /> \frac{sin \pi}{\pi} = 0<br />
But...
<br /> cos \pi = -1<br />
Can You Solve These Challenging Calculus Problems?
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The discussion centers around finding and developing 15 slightly difficult calculus problems for a course project, specifically for first and second-year calculus. Participants suggest focusing on topics such as trigonometric identities and problems involving limits, particularly the limit of sin(x)/x as x approaches 0. Various methods for proving this limit are discussed, including geometric interpretations and L'Hôpital's rule. Additionally, users share resources like tutorials from Harvey Mudd College and propose specific problems, including integrals and optimization challenges. The conversation highlights the collaborative effort to enhance understanding of calculus concepts.
What? Take x = pi.
<br /> \frac{sin \pi}{\pi} = 0<br />
But...
<br /> cos \pi = -1<br />
- #32
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Let t = time when this question was asked.
Find t.
Find t.
- #33
weagle2008
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l'Hôpital 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.
- #34
Anonymous217
- 355
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weagle2008 said:
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.
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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