Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-2+x^2-x)/(-2+x+3*x^2)
-
Limit of tan(-1+x)/(-1+x^2)
-
Limit of (-1+sin(x))/cos(x)
-
Limit of sin(5*x)/sin(4*x)
- Integral of d{x}:
- sin(5*x)/sin(4*x)
-
Identical expressions
- sin(five *x)/sin(four *x)
- sinus of (5 multiply by x) divide by sinus of (4 multiply by x)
- sinus of (five multiply by x) divide by sinus of (four multiply by x)
- sin(5x)/sin(4x)
- sin5x/sin4x
- sin(5*x) divide by sin(4*x)
Limit of the function sin(5*x)/sin(4*x)
The solution
You have entered
[src]
/sin(5*x)\ lim |--------| x->0+\sin(4*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right)$$
Limit(sin(5*x)/sin(4*x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{x} \frac{x}{\sin{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(4 x \right)}}\right)$$
=
Do replacement
$$u = 5 x$$
and
$$v = 4 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(4 x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{5 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{4 \sin{\left(v \right)}}\right)$$
=
$$\frac{5 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)}{4}$$
=
$$\frac{5 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{4}$$
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
=
$$\frac{5 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{4}$$
=
$$\frac{5}{4}$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right) = \frac{5}{4}$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{x} \frac{x}{\sin{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(4 x \right)}}\right)$$
=
Do replacement
$$u = 5 x$$
and
$$v = 4 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(4 x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{5 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{4 \sin{\left(v \right)}}\right)$$
=
$$\frac{5 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)}{4}$$
=
$$\frac{5 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{4}$$
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
=
$$\frac{5 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{4}$$
=
$$\frac{5}{4}$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right) = \frac{5}{4}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(5 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(4 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(5 x \right)}}{\frac{d}{d x} \sin{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5 \cos{\left(5 x \right)}}{4 \cos{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+} \frac{5}{4}$$
=
$$\lim_{x \to 0^+} \frac{5}{4}$$
=
$$\frac{5}{4}$$
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
$$\lim_{x \to 0^+} \sin{\left(5 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(4 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(5 x \right)}}{\frac{d}{d x} \sin{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5 \cos{\left(5 x \right)}}{4 \cos{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+} \frac{5}{4}$$
=
$$\lim_{x \to 0^+} \frac{5}{4}$$
=
$$\frac{5}{4}$$
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(5*x)\ lim |--------| x->0+\sin(4*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right)$$
5/4
$$\frac{5}{4}$$
= 1.25
/sin(5*x)\ lim |--------| x->0-\sin(4*x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(5 x \right)}}{\sin{\left(4 x \right)}}\right)$$
5/4
$$\frac{5}{4}$$
= 1.25
= 1.25
The graph