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sin(10*x)/tan(5*x)
Limit of the function/ sin(10*x)/tan(5*x)

Limit of the function sin(10*x)/tan(5*x)

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     /sin(10*x)\
 lim |---------|
x->0+\ tan(5*x)/
lim⁡x→0+(sin⁡(10x)tan⁡(5x))\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) 
Limit(sin(10*x)/tan(5*x), x, 0)
Detail solution
Let's take the limit
lim⁡x→0+(sin⁡(10x)tan⁡(5x))\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)
transform
lim⁡x→0+(sin⁡(10x)tan⁡(5x))=lim⁡x→0+(sin⁡(10x)xxtan⁡(5x))\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{x} \frac{x}{\tan{\left(5 x \right)}}\right)
=
lim⁡x→0+(sin⁡(10x)x)lim⁡x→0+(xtan⁡(5x))\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(5 x \right)}}\right)
=
Do replacement
u=10xu = 10 x
and
v=5xv = 5 x
then
lim⁡x→0+(sin⁡(10x)tan⁡(5x))=lim⁡u→0+(10sin⁡(u)u)lim⁡v→0+(v5tan⁡(v))\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{10 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{5 \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⁡(10x)tan⁡(5x))=2\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 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⁡(10x)=0\lim_{x \to 0^+} \sin{\left(10 x \right)} = 0
and limit for the denominator is
lim⁡x→0+tan⁡(5x)=0\lim_{x \to 0^+} \tan{\left(5 x \right)} = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(sin⁡(10x)tan⁡(5x))\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)
=
lim⁡x→0+(ddxsin⁡(10x)ddxtan⁡(5x))\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(10 x \right)}}{\frac{d}{d x} \tan{\left(5 x \right)}}\right)
=
lim⁡x→0+(10cos⁡(10x)5tan⁡2(5x)+5)\lim_{x \to 0^+}\left(\frac{10 \cos{\left(10 x \right)}}{5 \tan^{2}{\left(5 x \right)} + 5}\right)
=
lim⁡x→0+(105tan⁡2(5x)+5)\lim_{x \to 0^+}\left(\frac{10}{5 \tan^{2}{\left(5 x \right)} + 5}\right)
=
lim⁡x→0+(105tan⁡2(5x)+5)\lim_{x \to 0^+}\left(\frac{10}{5 \tan^{2}{\left(5 x \right)} + 5}\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
Other limits x→0, -oo, +oo, 1
lim⁡x→0−(sin⁡(10x)tan⁡(5x))=2\lim_{x \to 0^-}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = 2
More at x→0 from the left
lim⁡x→0+(sin⁡(10x)tan⁡(5x))=2\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = 2
lim⁡x→∞(sin⁡(10x)tan⁡(5x))\lim_{x \to \infty}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)
More at x→oo
lim⁡x→1−(sin⁡(10x)tan⁡(5x))=sin⁡(10)tan⁡(5)\lim_{x \to 1^-}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \frac{\sin{\left(10 \right)}}{\tan{\left(5 \right)}}
More at x→1 from the left
lim⁡x→1+(sin⁡(10x)tan⁡(5x))=sin⁡(10)tan⁡(5)\lim_{x \to 1^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \frac{\sin{\left(10 \right)}}{\tan{\left(5 \right)}}
More at x→1 from the right
lim⁡x→−∞(sin⁡(10x)tan⁡(5x))\lim_{x \to -\infty}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)
More at x→-oo
One‐sided limits [src] 
     /sin(10*x)\
 lim |---------|
x->0+\ tan(5*x)/
lim⁡x→0+(sin⁡(10x)tan⁡(5x))\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) 
2
22 
= 2.0
     /sin(10*x)\
 lim |---------|
x->0-\ tan(5*x)/
lim⁡x→0−(sin⁡(10x)tan⁡(5x))\lim_{x \to 0^-}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) 
2
22 
= 2.0
= 2.0
Numerical answer [src] 
2.0
2.0
The graph
Limit of the function sin(10*x)/tan(5*x)
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