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

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

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     /sin(2*x)\
 lim |--------|
x->0+\ tan(x) /
lim⁡x→0+(sin⁡(2x)tan⁡(x))\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) 
Limit(sin(2*x)/tan(x), x, 0)
Detail solution
Let's take the limit
lim⁡x→0+(sin⁡(2x)tan⁡(x))\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)
transform
lim⁡x→0+(sin⁡(2x)tan⁡(x))=lim⁡x→0+(sin⁡(2x)xxtan⁡(x))\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x} \frac{x}{\tan{\left(x \right)}}\right)
=
lim⁡x→0+(sin⁡(2x)x)lim⁡x→0+(xtan⁡(x))\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(x \right)}}\right)
=
Do replacement
u=2xu = 2 x
and
v=xv = x
then
lim⁡x→0+(sin⁡(2x)tan⁡(x))=lim⁡u→0+(2sin⁡(u)u)lim⁡v→0+(vtan⁡(v))\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{2 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)
=
2lim⁡u→0+(sin⁡(u)u)lim⁡v→0+(vtan⁡(v))2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)
=
2lim⁡u→0+(sin⁡(u)u)(lim⁡v→0+(tan⁡(v)v))−12 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}
transform
lim⁡v→0+(tan⁡(v)v)=lim⁡v→0+(sin⁡(v)vcos⁡(v))\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right) = \lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v \cos{\left(v \right)}}\right)
=
lim⁡v→0+(sin⁡(v)v)lim⁡v→0+1cos⁡(v)=lim⁡v→0+(sin⁡(v)v)\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right) \lim_{v \to 0^+} \frac{1}{\cos{\left(v \right)}} = \lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)
The limit
lim⁡u→0+(sin⁡(u)u)\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)
is first remarkable limit, is equal to 1.

The final answer:
lim⁡x→0+(sin⁡(2x)tan⁡(x))=2\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(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⁡(2x)=0\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0
and limit for the denominator is
lim⁡x→0+tan⁡(x)=0\lim_{x \to 0^+} \tan{\left(x \right)} = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(sin⁡(2x)tan⁡(x))\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)
=
lim⁡x→0+(ddxsin⁡(2x)ddxtan⁡(x))\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(2 x \right)}}{\frac{d}{d x} \tan{\left(x \right)}}\right)
=
lim⁡x→0+(2cos⁡(2x)tan⁡2(x)+1)\lim_{x \to 0^+}\left(\frac{2 \cos{\left(2 x \right)}}{\tan^{2}{\left(x \right)} + 1}\right)
=
lim⁡x→0+(2tan⁡2(x)+1)\lim_{x \to 0^+}\left(\frac{2}{\tan^{2}{\left(x \right)} + 1}\right)
=
lim⁡x→0+(2tan⁡2(x)+1)\lim_{x \to 0^+}\left(\frac{2}{\tan^{2}{\left(x \right)} + 1}\right)
=
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
02468-8-6-4-2-101004
Plot the graph
Rapid solution [src] 
2
22
Expand and simplify
One‐sided limits [src] 
     /sin(2*x)\
 lim |--------|
x->0+\ tan(x) /
lim⁡x→0+(sin⁡(2x)tan⁡(x))\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) 
2
22 
= 2.0
     /sin(2*x)\
 lim |--------|
x->0-\ tan(x) /
lim⁡x→0−(sin⁡(2x)tan⁡(x))\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) 
2
22 
= 2.0
= 2.0
Other limits x→0, -oo, +oo, 1
lim⁡x→0−(sin⁡(2x)tan⁡(x))=2\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = 2
More at x→0 from the left
lim⁡x→0+(sin⁡(2x)tan⁡(x))=2\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = 2
lim⁡x→∞(sin⁡(2x)tan⁡(x))\lim_{x \to \infty}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)
More at x→oo
lim⁡x→1−(sin⁡(2x)tan⁡(x))=sin⁡(2)tan⁡(1)\lim_{x \to 1^-}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \frac{\sin{\left(2 \right)}}{\tan{\left(1 \right)}}
More at x→1 from the left
lim⁡x→1+(sin⁡(2x)tan⁡(x))=sin⁡(2)tan⁡(1)\lim_{x \to 1^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \frac{\sin{\left(2 \right)}}{\tan{\left(1 \right)}}
More at x→1 from the right
lim⁡x→−∞(sin⁡(2x)tan⁡(x))\lim_{x \to -\infty}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)
More at x→-oo
Numerical answer [src] 
2.0
2.0
The graph
Limit of the function sin(2*x)/tan(x)
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