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    Hi there!

    Just a question regarding cot x. I was tackling a question on integration for C4 and came across needing to substitute pi/2 into 1/cotx. I didn't think this was possible as tan(pi/2) is undefined, so I thought no solution.

    However, the mark scheme has that pi/2 into 1/cotx gives 0. Is this because 1/undefined = so tiny it is zero? Or is the mark scheme wrong?

    Thanks very much for any help
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    (Original post by Mynameisnotearl)
    Hi there!

    Just a question regarding cot x. I was tackling a question on integration for C4 and came across needing to substitute pi/2 into 1/cotx. I didn't think this was possible as tan(pi/2) is undefined, so I thought no solution.

    However, the mark scheme has that pi/2 into 1/cotx gives 0. Is this because 1/undefined = so tiny it is zero? Or is the mark scheme wrong?

    Thanks very much for any help
    \displaystyle \cot x = \dfrac{1}{\tan x} = \dfrac{\cos x }{\sin x}

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    (Original post by Mynameisnotearl)
    Hi there!

    Just a question regarding cot x. I was tackling a question on integration for C4 and came across needing to substitute pi/2 into 1/cotx. I didn't think this was possible as tan(pi/2) is undefined, so I thought no solution.

    However, the mark scheme has that pi/2 into 1/cotx gives 0. Is this because 1/undefined = so tiny it is zero? Or is the mark scheme wrong?

    Thanks very much for any help
    Well \tan(x) = \dfrac{\sin(x)}{\cos(x)} and at x= \dfrac{\pi}{2} you end up with a 0 on the denominator.

    \cot(x) = \dfrac{1}{\tan(x)} = \dfrac{\cos(x)}{\sin(x)} at x= \dfrac{\pi}{2} you do not end up with a 0 on the denominator.

    Also, if a function f(x) \rightarrow \infty at some x \rightarrow a then \dfrac{1}{f(x)} \rightarrow 0 at x \rightarrow a
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    ^^ They need to put some analysis in the A-level syllabus.
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    Thanks everyone

    Seems obvious and simple now - kicking myself for spending 20 mins stuck on this.
 
 
 
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