Trigonometric Derivative as a Difference Quotient

Question

I understand that the 2x might be treated as h. However, I don't know how to deal with cotangent.

Limit (as h -> 0) of [8 * cot³(2x + pi/3) - 8 * cot³(pi/3)]/h

Answer 1

Limit (as h -> 0) of [8 * cot³(2x + pi/3) - 8 * cot³(pi/3)]/h

letting x=h, It is hard not to use calculus because the limit is 8*d(cot³(2x))dx evaluated at pi/6

which with power rule is 8*3*(cot²(2x))*d[cot(2x)]/dx evaluated at x=pi/6

which with chain rule is 8*3*cot²(2x)*(d[cot(2x)]/d[2x])*(d(2x)/dx)) evaluated at pi/6

substituting z=2x, need to know d(cot(z))/dz

=lim h approaches zero [cot(z+h)-cot(z)]/h

=lim h->0 [(cot(z)-tan(h))/(1+cot(z)tan(h))-cot(z)]/h

=lim h->0 [(cot(z)-tan(h))-cot(z)-cot²(z)*tan(h)]/[h*(1+cot(z)tan(h))]

notice h*tan(h)*cos(z) is zero when h->0 for any z so denominator limit is h*1

numerator is -tan(h)-cot²(z)*tan(h)

d(cot(z))/dz=lim h->0 [-tan(h)-cot²(z)*tan(h)]/h

can prove lim h->0 (sin(h))/h =1,,,,,, of course lim h->0 cos(h)=1

d(cot(z))dz = -1-cot²(z)=-csc²(z)=-csc²(2x)

at x=pi/6 the oriiginal limit d(cot³(2x))/dx is 8*3*cot²(pi/3)*(-)csc²(pi/3)*2