Mister Exam
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How to use it?
- Limit of the function:
-
Limit of sin(3*x)/sin(x)
-
Limit of n/factorial(n)
-
Limit of tan(3*x)
-
Limit of sqrt(3)/3
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Identical expressions
- sin(three *x)/sin(x)
- sinus of (3 multiply by x) divide by sinus of (x)
- sinus of (three multiply by x) divide by sinus of (x)
- sin(3x)/sin(x)
- sin3x/sinx
- sin(3*x) divide by sin(x)
-
Similar expressions
- sin(3*x)/sinx
Limit of the function sin(3*x)/sin(x)
The solution
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/sin(3*x)\ lim |--------| x->0+\ sin(x) /
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right)$$
Limit(sin(3*x)/sin(x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x} \frac{x}{\sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(x \right)}}\right)$$
=
Do replacement
$$u = 3 x$$
and
$$v = x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{3 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)$$
=
$$3 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)$$
=
$$3 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}$$
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
=
$$3 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}$$
=
$$3$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = 3$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x} \frac{x}{\sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(x \right)}}\right)$$
=
Do replacement
$$u = 3 x$$
and
$$v = x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{3 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)$$
=
$$3 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)$$
=
$$3 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}$$
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
=
$$3 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}$$
=
$$3$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = 3$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(3 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(3 x \right)}}{\frac{d}{d x} \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3 \cos{\left(3 x \right)}}{\cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3}{\cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3}{\cos{\left(x \right)}}\right)$$
=
$$3$$
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
$$\lim_{x \to 0^+} \sin{\left(3 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(3 x \right)}}{\frac{d}{d x} \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3 \cos{\left(3 x \right)}}{\cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3}{\cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3}{\cos{\left(x \right)}}\right)$$
=
$$3$$
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(3*x)\ lim |--------| x->0+\ sin(x) /
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right)$$
3
$$3$$
= 3
/sin(3*x)\ lim |--------| x->0-\ sin(x) /
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right)$$
3
$$3$$
= 3
= 3
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = 3$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = 3$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \left\langle -\infty, \infty\right\rangle$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \frac{\sin{\left(3 \right)}}{\sin{\left(1 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \frac{\sin{\left(3 \right)}}{\sin{\left(1 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \left\langle -\infty, \infty\right\rangle$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = 3$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \left\langle -\infty, \infty\right\rangle$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \frac{\sin{\left(3 \right)}}{\sin{\left(1 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \frac{\sin{\left(3 \right)}}{\sin{\left(1 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(3 x \right)}}{\sin{\left(x \right)}}\right) = \left\langle -\infty, \infty\right\rangle$$
More at x→-oo
The graph