How do you find the limit of # (acota)/(sina) # as a approaches 0?

Question

1074 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T06:34:08

Rewriting with #cota=cosa/sina#:

#L=lim_(ararr0)(acota)/sina=lim_(ararr0)(acosa)/sin^2a#

This is in the indeterminate form #0/0#, so we can apply L'Hopital's rule and take the derivatives of the numerator and denominator individually:

#L=lim_(ararr0)(d/(da)(acosa))/(d/(da)sin^2a)=lim_(ararr0)(cosa-asina)/(2sinacosa)#

This is now no longer an indeterminate form, since it is the form #1/0#. Thus, the limit does not exist (there's an asymptote at #a=0#).

Answer 2 · 2016-11-09T06:52:58

Non-existing, #+-oo#, disclosing antipodal-infinite discontinuity.

#(a cot a)/sin a#=(a/sin a)(cos a/sin a)=(cos a)(csc a)

The are two approaches # to 0, 0_ + and 0_ _ #

#lim a to 0_ + csc a =oo and lim a to 0_ _ csc a = -oo#

But, either way, both (a /sina) and cos a to 1.

So, the limit of the product = product of the limits

#=(1)(1)(+-oo)# and can be symbolized as non-exiting

#+-oo# that reveals antipodal-infinite discontinuity at a = 0..