Limit of (1/x - 1/sin(x)) as x approaches zero

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I work out the limit of (1/x - 1/sin(x)) as x approaches zero. This is an exercise in the book by Michael Spivak titled Calculus.

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My original approach would have been, since x approaches 0, sinx approaches x (small angle approximation), so lim x -> 0 (1/x - 1/sinx) = lim x -> 0 (1/x - 1/x) = lim x -> 0 (0) = 0

juliussoldan

Since all of my teachers have never accepted l'Hopital as a solving method, I would've divided both the numerator and denominator by x, since sinx/x tends to 1 when x tends to 0, then it becomes (1-1)/1 = 0. A trick

imadchafik

Ha, I wonder what the record is for iteratively applying L'Hôpital's rule to determine the limit!

Enigma

Thanks for once again making Spivak's book feel less intimidating and more inviting! 👏 ❤️

surrealistidealist

Best of the Best! tank you! You save me!

leonardgomesgoncalves

You should check out calculus with analytic geometry by Robert Ellis and Denny Gulick it is a good calculus book

wendysofficialyoutube

I just got the answer 0 with a legit and short solution ...if my solution is right, then i don't think so much trignometry is needed heere

bhavishyasharma

Since Sin(x)~x as x->0, therefore lim 1/x-1/sinx is 0

karimelamili

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