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sin(4*x)/sin(2*x)
Limit of the function/ sin(4*x)/sin(2*x)

Limit of the function sin(4*x)/sin(2*x)

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     /sin(4*x)\
 lim |--------|
x->0+\sin(2*x)/
lim⁡x→0+(sin⁡(4x)sin⁡(2x))\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) 
Limit(sin(4*x)/sin(2*x), x, 0)
Detail solution
Let's take the limit
lim⁡x→0+(sin⁡(4x)sin⁡(2x))\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)
transform
lim⁡x→0+(sin⁡(4x)sin⁡(2x))=lim⁡x→0+(sin⁡(4x)xxsin⁡(2x))\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x} \frac{x}{\sin{\left(2 x \right)}}\right)
=
lim⁡x→0+(sin⁡(4x)x)lim⁡x→0+(xsin⁡(2x))\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(2 x \right)}}\right)
=
Do replacement
u=4xu = 4 x
and
v=2xv = 2 x
then
lim⁡x→0+(sin⁡(4x)sin⁡(2x))=lim⁡x→0+(sin⁡(4x)x)lim⁡x→0+(xsin⁡(2x))\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(2 x \right)}}\right)
lim⁡x→0+(sin⁡(4x)sin⁡(2x))=lim⁡u→0+(4sin⁡(u)u)lim⁡v→0+(v2sin⁡(v))\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{4 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{2 \sin{\left(v \right)}}\right)
=
2lim⁡u→0+(sin⁡(u)u)lim⁡v→0+(vsin⁡(v))2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)
=
2lim⁡u→0+(sin⁡(u)u)(lim⁡v→0+(sin⁡(v)v))−12 \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}
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
=
2(lim⁡v→0+(sin⁡(v)v))−12 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}
=
22

The final answer:
lim⁡x→0+(sin⁡(4x)sin⁡(2x))=2\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = 2
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
lim⁡x→0+sin⁡(4x)=0\lim_{x \to 0^+} \sin{\left(4 x \right)} = 0
and limit for the denominator is
lim⁡x→0+sin⁡(2x)=0\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(sin⁡(4x)sin⁡(2x))\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)
=
lim⁡x→0+(ddxsin⁡(4x)ddxsin⁡(2x))\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(4 x \right)}}{\frac{d}{d x} \sin{\left(2 x \right)}}\right)
=
lim⁡x→0+(2cos⁡(4x)cos⁡(2x))\lim_{x \to 0^+}\left(\frac{2 \cos{\left(4 x \right)}}{\cos{\left(2 x \right)}}\right)
=
lim⁡x→0+2\lim_{x \to 0^+} 2
=
lim⁡x→0+2\lim_{x \to 0^+} 2
=
22
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)
The graph
02468-8-6-4-2-10105-5
Plot the graph
One‐sided limits [src] 
     /sin(4*x)\
 lim |--------|
x->0+\sin(2*x)/
lim⁡x→0+(sin⁡(4x)sin⁡(2x))\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) 
2
22 
= 2.0
     /sin(4*x)\
 lim |--------|
x->0-\sin(2*x)/
lim⁡x→0−(sin⁡(4x)sin⁡(2x))\lim_{x \to 0^-}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) 
2
22 
= 2.0
= 2.0
Rapid solution [src] 
2
22
Expand and simplify
Numerical answer [src] 
2.0
2.0
The graph
Limit of the function sin(4*x)/sin(2*x)
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