Question

limit tan m theta by tan n theta and x extends to 0​

Answer 1

I am using A in place of theta

if we put A = 0 we get

tan m0/ sin n0

0/0 ( indeterminate)

so we can use L'Hospital Rule here

let f(A) = tan mA

g(A) = sin nA

with respective limit as per L'Hospital rule

lim f(A) / g(A). = lim f'(A) / g'(A)

f'(A) = d tan mA/dA = msec^2 mA

g'(A) = d sin nA/dA = n cos nA

our limit becomes

lim m sec^2 mA / n cos nA

= m sec^2 0 / n cos 0

= m*1 / n*1

= m/n