Limit with trigonometric functions

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The limit of Tanx/(1-Cosx) as x approaches π from the left evaluates to 0, indicating that it does not equal infinity. The second limit, Sec2x/Sinx, also approaches 0 as x approaches π, confirming that both limits are equal and finite. The discussion highlights that L'Hopital's Rule is not applicable since the limits do not present indeterminate forms. The conclusion is that the statement claiming the limits equal infinity is false. Thus, the original assertion is incorrect based on the evaluations provided.
Jules18
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Homework Statement



True or false?

lim Tanx/(1-Cosx) = lim Sec2x/Sinx = Infinity

(limits are as x approaches \pi from the left)

The Attempt at a Solution



I tried just plugging \pi into x in the first limit, and I ended up getting 0/2, which exists but is just 0. So I said false. Am I simplifying it too much? because what I did seems to make sense but it looks like that's not what they wanted me to do.
 
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Isn't it just 0/2=0?
 


The original limit was neither of the indeterminate forms [0/0] or [infinity/infinity], so L'Hopital's Rule doesn't apply here. Your first approach was correct.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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