Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (-6+x^2-x)/(6+x^2-5*x)
-
Limit of log(1+x^2)/(-e^(-x)+cos(3*x))
-
Limit of -2+|-2+x|/x
-
Limit of (-4-8*x+8*x^2)/(3+8*x)
- Integral of d{x}:
-
sin(4*x)/sin(2*x)
-
Identical expressions
- sin(four *x)/sin(two *x)
- sinus of (4 multiply by x) divide by sinus of (2 multiply by x)
- sinus of (four multiply by x) divide by sinus of (two multiply by x)
- sin(4x)/sin(2x)
- sin4x/sin2x
- sin(4*x) divide by sin(2*x)
Limit of the function sin(4*x)/sin(2*x)
The solution
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/sin(4*x)\ lim |--------| x->0+\sin(2*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)$$
Limit(sin(4*x)/sin(2*x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x} \frac{x}{\sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(2 x \right)}}\right)$$
=
Do replacement
$$u = 4 x$$
and
$$v = 2 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(2 x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{4 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{2 \sin{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
and
$$\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)$$
is first remarkable limit, is equal to 1.
then
=
$$2 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}$$
=
$$2$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = 2$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x} \frac{x}{\sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(2 x \right)}}\right)$$
=
Do replacement
$$u = 4 x$$
and
$$v = 2 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(2 x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{4 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{2 \sin{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
and
$$\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)$$
is first remarkable limit, is equal to 1.
then
=
$$2 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}$$
=
$$2$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right) = 2$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(4 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(4 x \right)}}{\frac{d}{d x} \sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \cos{\left(4 x \right)}}{\cos{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+} 2$$
=
$$\lim_{x \to 0^+} 2$$
=
$$2$$
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)
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(4 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(4 x \right)}}{\frac{d}{d x} \sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \cos{\left(4 x \right)}}{\cos{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+} 2$$
=
$$\lim_{x \to 0^+} 2$$
=
$$2$$
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
One‐sided limits
[src]
/sin(4*x)\ lim |--------| x->0+\sin(2*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)$$
2
$$2$$
= 2.0
/sin(4*x)\ lim |--------| x->0-\sin(2*x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(4 x \right)}}{\sin{\left(2 x \right)}}\right)$$
2
$$2$$
= 2.0
= 2.0
The graph