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

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

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     /sin(m*x)\
 lim |--------|
x->0+\sin(n*x)/
lim⁡x→0+(sin⁡(mx)sin⁡(nx))\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) 
Limit(sin(m*x)/sin(n*x), x, 0)
Detail solution
Let's take the limit
lim⁡x→0+(sin⁡(mx)sin⁡(nx))\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)
transform
lim⁡x→0+(sin⁡(mx)sin⁡(nx))=lim⁡x→0+(sin⁡(mx)xxsin⁡(nx))\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x} \frac{x}{\sin{\left(n x \right)}}\right)
=
lim⁡x→0+(sin⁡(mx)x)lim⁡x→0+(xsin⁡(nx))\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(n x \right)}}\right)
=
Do replacement
u=mxu = m x
and
v=nxv = n x
then
lim⁡x→0+(sin⁡(mx)sin⁡(nx))=lim⁡x→0+(sin⁡(mx)x)lim⁡x→0+(xsin⁡(nx))\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(n x \right)}}\right)
lim⁡x→0+(sin⁡(mx)sin⁡(nx))=lim⁡u→0+(msin⁡(u)u)lim⁡v→0+(vnsin⁡(v))\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{m \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{n \sin{\left(v \right)}}\right)
=
mlim⁡u→0+(sin⁡(u)u)lim⁡v→0+(vsin⁡(v))n\frac{m \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)}{n}
=
mlim⁡u→0+(sin⁡(u)u)(lim⁡v→0+(sin⁡(v)v))−1n\frac{m \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{n}
The limit
lim⁡u→0+(sin⁡(u)u)\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)
and
lim⁡v→0+(sin⁡(v)v)\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)
is first remarkable limit, is equal to 1.
then
=
m(lim⁡v→0+(sin⁡(v)v))−1n\frac{m \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{n}
=
mn\frac{m}{n}

The final answer:
lim⁡x→0+(sin⁡(mx)sin⁡(nx))=mn\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \frac{m}{n}
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
lim⁡x→0+sin⁡(mx)=0\lim_{x \to 0^+} \sin{\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+(sin⁡(mx)sin⁡(nx))\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)
=
lim⁡x→0+(∂∂xsin⁡(mx)∂∂xsin⁡(nx))\lim_{x \to 0^+}\left(\frac{\frac{\partial}{\partial x} \sin{\left(m x \right)}}{\frac{\partial}{\partial x} \sin{\left(n x \right)}}\right)
=
lim⁡x→0+(mcos⁡(mx)ncos⁡(nx))\lim_{x \to 0^+}\left(\frac{m \cos{\left(m x \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] 
     /sin(m*x)\
 lim |--------|
x->0+\sin(n*x)/
lim⁡x→0+(sin⁡(mx)sin⁡(nx))\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) 
m
-
n
mn\frac{m}{n} 
     /sin(m*x)\
 lim |--------|
x->0-\sin(n*x)/
lim⁡x→0−(sin⁡(mx)sin⁡(nx))\lim_{x \to 0^-}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) 
m
-
n
mn\frac{m}{n} 
m/n
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