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

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

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     /sin(5*x)\
 lim |--------|
x->0+\  2*x   /
lim⁡x→0+(sin⁡(5x)2x)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) 
Limit(sin(5*x)/((2*x)), x, 0)
Detail solution
Let's take the limit
lim⁡x→0+(sin⁡(5x)2x)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right)
Do replacement
u=5xu = 5 x
then
lim⁡x→0+(sin⁡(5x)2x)=lim⁡u→0+(5sin⁡(u)2u)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) = \lim_{u \to 0^+}\left(\frac{5 \sin{\left(u \right)}}{2 u}\right)
=
5lim⁡u→0+(sin⁡(u)u)2\frac{5 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{2}
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⁡(5x)2x)=52\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) = \frac{5}{2}
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
lim⁡x→0+sin⁡(5x)=0\lim_{x \to 0^+} \sin{\left(5 x \right)} = 0
and limit for the denominator is
lim⁡x→0+(2x)=0\lim_{x \to 0^+}\left(2 x\right) = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(sin⁡(5x)2x)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right)
=
Let's transform the function under the limit a few
lim⁡x→0+(sin⁡(5x)2x)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right)
=
lim⁡x→0+(ddxsin⁡(5x)ddx2x)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(5 x \right)}}{\frac{d}{d x} 2 x}\right)
=
lim⁡x→0+(5cos⁡(5x)2)\lim_{x \to 0^+}\left(\frac{5 \cos{\left(5 x \right)}}{2}\right)
=
lim⁡x→0+52\lim_{x \to 0^+} \frac{5}{2}
=
lim⁡x→0+52\lim_{x \to 0^+} \frac{5}{2}
=
52\frac{5}{2}
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(5*x)\
 lim |--------|
x->0+\  2*x   /
lim⁡x→0+(sin⁡(5x)2x)\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) 
5/2
52\frac{5}{2} 
= 2.5
     /sin(5*x)\
 lim |--------|
x->0-\  2*x   /
lim⁡x→0−(sin⁡(5x)2x)\lim_{x \to 0^-}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) 
5/2
52\frac{5}{2} 
= 2.5
= 2.5
Rapid solution [src] 
5/2
52\frac{5}{2}
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→0−(sin⁡(5x)2x)=52\lim_{x \to 0^-}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) = \frac{5}{2}
More at x→0 from the left
lim⁡x→0+(sin⁡(5x)2x)=52\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) = \frac{5}{2}
lim⁡x→∞(sin⁡(5x)2x)=0\lim_{x \to \infty}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) = 0
More at x→oo
lim⁡x→1−(sin⁡(5x)2x)=sin⁡(5)2\lim_{x \to 1^-}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) = \frac{\sin{\left(5 \right)}}{2}
More at x→1 from the left
lim⁡x→1+(sin⁡(5x)2x)=sin⁡(5)2\lim_{x \to 1^+}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) = \frac{\sin{\left(5 \right)}}{2}
More at x→1 from the right
lim⁡x→−∞(sin⁡(5x)2x)=0\lim_{x \to -\infty}\left(\frac{\sin{\left(5 x \right)}}{2 x}\right) = 0
More at x→-oo
Numerical answer [src] 
2.5
2.5
The graph
Limit of the function sin(5*x)/(2*x)
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