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

Question

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

Elijah Abram · Accepted Answer

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#).

Ezekiel Alexander · Answer

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..

Luke Alvarez · Answer

undefined To find the limit of (acot(a))/(sin(a)) as a approaches 0, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately and then take the limit as a approaches 0.

Differentiating the numerator, we get -acsc^2(a), and differentiating the denominator, we get cos(a).

Taking the limit as a approaches 0, we have (-acsc^2(a))/(cos(a)).

Substituting a = 0 into the expression, we get (-0csc^2(0))/(cos(0)).

Since csc(0) is undefined, we cannot directly substitute a = 0. However, we can simplify the expression further.

Using the trigonometric identity csc^2(a) = 1 + cot^2(a), we can rewrite the expression as (-a(1 + cot^2(a)))/(cos(a)).

Taking the limit as a approaches 0, we have (-0(1 + cot^2(0)))/(cos(0)).

Simplifying this expression, we get (0)/(1), which equals 0.

Therefore, the limit of (acot(a))/(sin(a)) as a approaches 0 is 0.