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sin(2*x)/x

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

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     /sin(2*x)\
 lim |--------|
x->0+\   x    /
limx0+(sin(2x)x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right)
Limit(sin(2*x)/x, x, 0)
Detail solution
Let's take the limit
limx0+(sin(2x)x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right)
Do replacement
u=2xu = 2 x
then
limx0+(sin(2x)x)=limu0+(2sin(u)u)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right) = \lim_{u \to 0^+}\left(\frac{2 \sin{\left(u \right)}}{u}\right)
=
2limu0+(sin(u)u)2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)
The limit
limu0+(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:
limx0+(sin(2x)x)=2\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right) = 2
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
limx0+sin(2x)=0\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0
and limit for the denominator is
limx0+x=0\lim_{x \to 0^+} x = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
limx0+(sin(2x)x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right)
=
limx0+(ddxsin(2x)ddxx)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(2 x \right)}}{\frac{d}{d x} x}\right)
=
limx0+(2cos(2x))\lim_{x \to 0^+}\left(2 \cos{\left(2 x \right)}\right)
=
limx0+2\lim_{x \to 0^+} 2
=
limx0+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
02468-8-6-4-2-10102.5-2.5
One‐sided limits [src]
     /sin(2*x)\
 lim |--------|
x->0+\   x    /
limx0+(sin(2x)x)\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right)
2
22
= 2.0
     /sin(2*x)\
 lim |--------|
x->0-\   x    /
limx0(sin(2x)x)\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{x}\right)
2
22
= 2.0
= 2.0
Other limits x→0, -oo, +oo, 1
limx0(sin(2x)x)=2\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{x}\right) = 2
More at x→0 from the left
limx0+(sin(2x)x)=2\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right) = 2
limx(sin(2x)x)=0\lim_{x \to \infty}\left(\frac{\sin{\left(2 x \right)}}{x}\right) = 0
More at x→oo
limx1(sin(2x)x)=sin(2)\lim_{x \to 1^-}\left(\frac{\sin{\left(2 x \right)}}{x}\right) = \sin{\left(2 \right)}
More at x→1 from the left
limx1+(sin(2x)x)=sin(2)\lim_{x \to 1^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right) = \sin{\left(2 \right)}
More at x→1 from the right
limx(sin(2x)x)=0\lim_{x \to -\infty}\left(\frac{\sin{\left(2 x \right)}}{x}\right) = 0
More at x→-oo
Rapid solution [src]
2
22
Numerical answer [src]
2.0
2.0
The graph
Limit of the function sin(2*x)/x