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Limit of the function/ tan(m*x)/sin(n*x)

Limit of the function tan(m*x)/sin(n*x)

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     /tan(m*x)\
 lim |--------|
x->0+\sin(n*x)/
lim⁡x→0+(tan⁡(mx)sin⁡(nx))\lim_{x \to 0^+}\left(\frac{\tan{\left(m x \right)}}{\sin{\left(n x \right)}}\right) 
Limit(tan(m*x)/sin(n*x), x, 0)
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
lim⁡x→0+tan⁡(mx)=0\lim_{x \to 0^+} \tan{\left(m x \right)} = 0
and limit for the denominator is
lim⁡x→0+sin⁡(nx)=0\lim_{x \to 0^+} \sin{\left(n x \right)} = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(tan⁡(mx)sin⁡(nx))\lim_{x \to 0^+}\left(\frac{\tan{\left(m x \right)}}{\sin{\left(n x \right)}}\right)
=
lim⁡x→0+(∂∂xtan⁡(mx)∂∂xsin⁡(nx))\lim_{x \to 0^+}\left(\frac{\frac{\partial}{\partial x} \tan{\left(m x \right)}}{\frac{\partial}{\partial x} \sin{\left(n x \right)}}\right)
=
lim⁡x→0+(m(tan⁡2(mx)+1)ncos⁡(nx))\lim_{x \to 0^+}\left(\frac{m \left(\tan^{2}{\left(m x \right)} + 1\right)}{n \cos{\left(n x \right)}}\right)
=
lim⁡x→0+(mn)\lim_{x \to 0^+}\left(\frac{m}{n}\right)
=
lim⁡x→0+(mn)\lim_{x \to 0^+}\left(\frac{m}{n}\right)
=
mn\frac{m}{n}
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
Rapid solution [src] 
m
-
n
mn\frac{m}{n}
Expand and simplify
One‐sided limits [src] 
     /tan(m*x)\
 lim |--------|
x->0+\sin(n*x)/
lim⁡x→0+(tan⁡(mx)sin⁡(nx))\lim_{x \to 0^+}\left(\frac{\tan{\left(m x \right)}}{\sin{\left(n x \right)}}\right) 
m
-
n
mn\frac{m}{n} 
     /tan(m*x)\
 lim |--------|
x->0-\sin(n*x)/
lim⁡x→0−(tan⁡(mx)sin⁡(nx))\lim_{x \to 0^-}\left(\frac{\tan{\left(m x \right)}}{\sin{\left(n x \right)}}\right) 
m
-
n
mn\frac{m}{n} 
m/n
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