Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (5+x-3*x^2)/(4-x+2*x^2)
-
Limit of ((3+4*x)/(-1+4*x))^(-3+2*x)
-
Limit of ((5+2*x)/(-1+2*x))^(3-x)
-
Limit of ((1+2*x)/(2+2*x))^x
-
Identical expressions
- sin(m*x)/sin(n*x)
- sinus of (m multiply by x) divide by sinus of (n multiply by x)
- sin(mx)/sin(nx)
- sinmx/sinnx
- sin(m*x) divide by sin(n*x)
Limit of the function sin(m*x)/sin(n*x)
The solution
You have entered
[src]
/sin(m*x)\ lim |--------| x->0+\sin(n*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)$$
Limit(sin(m*x)/sin(n*x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x} \frac{x}{\sin{\left(n x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(n x \right)}}\right)$$
=
Do replacement
$$u = m x$$
and
$$v = n x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(n x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{m \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{n \sin{\left(v \right)}}\right)$$
=
$$\frac{m \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)}{n}$$
=
$$\frac{m \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{n}$$
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
=
$$\frac{m \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{n}$$
=
$$\frac{m}{n}$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \frac{m}{n}$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x} \frac{x}{\sin{\left(n x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(n x \right)}}\right)$$
=
Do replacement
$$u = m x$$
and
$$v = n x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(n x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{m \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{n \sin{\left(v \right)}}\right)$$
=
$$\frac{m \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)}{n}$$
=
$$\frac{m \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{n}$$
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
=
$$\frac{m \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{n}$$
=
$$\frac{m}{n}$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right) = \frac{m}{n}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(m x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(n x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{\partial}{\partial x} \sin{\left(m x \right)}}{\frac{\partial}{\partial x} \sin{\left(n x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{m \cos{\left(m x \right)}}{n \cos{\left(n x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{m}{n}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{m}{n}\right)$$
=
$$\frac{m}{n}$$
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
$$\lim_{x \to 0^+} \sin{\left(m x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(n x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{\partial}{\partial x} \sin{\left(m x \right)}}{\frac{\partial}{\partial x} \sin{\left(n x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{m \cos{\left(m x \right)}}{n \cos{\left(n x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{m}{n}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{m}{n}\right)$$
=
$$\frac{m}{n}$$
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)
One‐sided limits
[src]
/sin(m*x)\ lim |--------| x->0+\sin(n*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)$$
m - n
$$\frac{m}{n}$$
/sin(m*x)\ lim |--------| x->0-\sin(n*x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(m x \right)}}{\sin{\left(n x \right)}}\right)$$
m - n
$$\frac{m}{n}$$
m/n