Find the limit LIM (x → π) sin 3x / Tan 5x and why? There is also a problem in the book solved to LIM (x → 0) (SEC ^ 2x TaNx) / x, (note that this is the square of secx) and then LIM (x → 0) TaNx / X. I don't know what the solution is. Continue with the lobita rule?

Find the limit LIM (x → π) sin 3x / Tan 5x and why? There is also a problem in the book solved to LIM (x → 0) (SEC ^ 2x TaNx) / x, (note that this is the square of secx) and then LIM (x → 0) TaNx / X. I don't know what the solution is. Continue with the lobita rule?

limsin3x/tan5x =lim3cos3x/[5(sec5x)^2] =(3/5)limcos3x(cos5x)^2 =(3/5)cos3π(cos5π)^2 =-3/5 limtanx/x=limx/x=1

When x approaches Pai Lim [sin (3x) / Tan (5x)]

lim【x→π】[sin(3x)]/[tan(5x)]
=lim【x→π】[-sin(3x-3π)]/[tan(5x-5π)]
=Lim [x → π] - (3x-3 π) / (5x-5 π) [Equivalent Infinitesimal Substitution]
=Lim [x → π] - 3 / 5
=-3/5

Given sin (x + 2Y) = 3sinx, find Tan (x + y) * Coty

sin(x+2y)=3sinx ,sin[(x+y)+y]=3sin[(x+y)-y] ,sin(x+y)cos(y)+cos(x+y)sin(y)=3[sin(x+y)cos(y)-cos(x+y)sin(y)] ,sin(x+y)cos(y)=2cos(x+y)sin(y) ,tan(x+y)*coty=2