Evaluate without L'Hopital: <munder> <mo movablelimits="true" form="prefix">lim <mrow cla

velinariojepvg

velinariojepvg

Answered question

2022-05-16

Evaluate without L'Hopital: lim x 3 ( x 3 ) csc π x
I find the indeterminate form of 0 or 0 0 . The latter tells me that L'Hopital's is an option, but since we haven't seen derivatives yet I'm not allowed to used it.
Previously I already tried swapping the csc π x for 1 sin π x but when doing this I can't seem to get rid of the sinus. I also believe that since the limit goes to 3 and not to 0, the lim x 0 sin a x a x = 1 rule is not an option either.
I tried making the sin ( π 2 ) so that =1, but without any success. Can anyone give me a hint?

Answer & Explanation

garcialdariamcy4q

garcialdariamcy4q

Beginner2022-05-17Added 15 answers

Hint: sin ( x ) = sin ( x + 3 π ). Try a substitution such that the limit you mentioned is an option!
hisyhauttaq84w

hisyhauttaq84w

Beginner2022-05-18Added 4 answers

By a change of variable,
lim x 3 x 3 sin π x = lim y 0 y sin ( π y + 3 π ) = 1 π lim y 0 π y sin ( π y ) .

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