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sin(2*x)/(5*x)
Limit of the function/ sin(2*x)/(5*x)

Limit of the function sin(2*x)/(5*x)

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     /sin(2*x)\
 lim |--------|
x->0+\  5*x   /
lim⁡x→0+(sin⁡(2x)5x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) 
Limit(sin(2*x)/((5*x)), x, 0)
Detail solution
Let's take the limit
lim⁡x→0+(sin⁡(2x)5x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right)
Do replacement
u=2xu = 2 x
then
lim⁡x→0+(sin⁡(2x)5x)=lim⁡u→0+(2sin⁡(u)5u)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) = \lim_{u \to 0^+}\left(\frac{2 \sin{\left(u \right)}}{5 u}\right)
=
2lim⁡u→0+(sin⁡(u)u)5\frac{2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{5}
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)5x)=25\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) = \frac{2}{5}
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+(5x)=0\lim_{x \to 0^+}\left(5 x\right) = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(sin⁡(2x)5x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right)
=
Let's transform the function under the limit a few
lim⁡x→0+(sin⁡(2x)5x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right)
=
lim⁡x→0+(ddxsin⁡(2x)ddx5x)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(2 x \right)}}{\frac{d}{d x} 5 x}\right)
=
lim⁡x→0+(2cos⁡(2x)5)\lim_{x \to 0^+}\left(\frac{2 \cos{\left(2 x \right)}}{5}\right)
=
lim⁡x→0+25\lim_{x \to 0^+} \frac{2}{5}
=
lim⁡x→0+25\lim_{x \to 0^+} \frac{2}{5}
=
25\frac{2}{5}
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
02468-8-6-4-2-10100.5-0.5
Plot the graph
Rapid solution [src] 
2/5
25\frac{2}{5}
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→0−(sin⁡(2x)5x)=25\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) = \frac{2}{5}
More at x→0 from the left
lim⁡x→0+(sin⁡(2x)5x)=25\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) = \frac{2}{5}
lim⁡x→∞(sin⁡(2x)5x)=0\lim_{x \to \infty}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) = 0
More at x→oo
lim⁡x→1−(sin⁡(2x)5x)=sin⁡(2)5\lim_{x \to 1^-}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) = \frac{\sin{\left(2 \right)}}{5}
More at x→1 from the left
lim⁡x→1+(sin⁡(2x)5x)=sin⁡(2)5\lim_{x \to 1^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) = \frac{\sin{\left(2 \right)}}{5}
More at x→1 from the right
lim⁡x→−∞(sin⁡(2x)5x)=0\lim_{x \to -\infty}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) = 0
More at x→-oo
One‐sided limits [src] 
     /sin(2*x)\
 lim |--------|
x->0+\  5*x   /
lim⁡x→0+(sin⁡(2x)5x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) 
2/5
25\frac{2}{5} 
= 0.4
     /sin(2*x)\
 lim |--------|
x->0-\  5*x   /
lim⁡x→0−(sin⁡(2x)5x)\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{5 x}\right) 
2/5
25\frac{2}{5} 
= 0.4
= 0.4
Numerical answer [src] 
0.4
0.4
The graph
Limit of the function sin(2*x)/(5*x)
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