Mister Exam
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How to use it?
- Limit of the function:
-
Limit of (-9+x^2)/(-3+x^2-2*x)
-
Limit of (-1+cos(x))/x^2
-
Limit of sin(17*x)/sin(12*x)
-
Limit of (5+x)/(-5+x^2+4*x)
-
Identical expressions
- sin(seventeen *x)/sin(twelve *x)
- sinus of (17 multiply by x) divide by sinus of (12 multiply by x)
- sinus of (seventeen multiply by x) divide by sinus of (twelve multiply by x)
- sin(17x)/sin(12x)
- sin17x/sin12x
- sin(17*x) divide by sin(12*x)
Limit of the function sin(17*x)/sin(12*x)
The solution
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[src]
/sin(17*x)\ lim |---------| x->0+\sin(12*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
Limit(sin(17*x)/sin(12*x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{x} \frac{x}{\sin{\left(12 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(12 x \right)}}\right)$$
=
Do replacement
$$u = 17 x$$
and
$$v = 12 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(12 x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{17 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{12 \sin{\left(v \right)}}\right)$$
=
$$\frac{17 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)}{12}$$
=
$$\frac{17 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{12}$$
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
=
$$\frac{17 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{12}$$
=
$$\frac{17}{12}$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{17}{12}$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{x} \frac{x}{\sin{\left(12 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(12 x \right)}}\right)$$
=
Do replacement
$$u = 17 x$$
and
$$v = 12 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\sin{\left(12 x \right)}}\right)$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{17 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{12 \sin{\left(v \right)}}\right)$$
=
$$\frac{17 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\sin{\left(v \right)}}\right)}{12}$$
=
$$\frac{17 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{12}$$
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
=
$$\frac{17 \left(\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)\right)^{-1}}{12}$$
=
$$\frac{17}{12}$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{17}{12}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(17 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(12 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(17 x \right)}}{\frac{d}{d x} \sin{\left(12 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{17 \cos{\left(17 x \right)}}{12 \cos{\left(12 x \right)}}\right)$$
=
$$\lim_{x \to 0^+} \frac{17}{12}$$
=
$$\lim_{x \to 0^+} \frac{17}{12}$$
=
$$\frac{17}{12}$$
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
$$\lim_{x \to 0^+} \sin{\left(17 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(12 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(17 x \right)}}{\frac{d}{d x} \sin{\left(12 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{17 \cos{\left(17 x \right)}}{12 \cos{\left(12 x \right)}}\right)$$
=
$$\lim_{x \to 0^+} \frac{17}{12}$$
=
$$\lim_{x \to 0^+} \frac{17}{12}$$
=
$$\frac{17}{12}$$
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(17*x)\ lim |---------| x->0+\sin(12*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
17 -- 12
$$\frac{17}{12}$$
= 1.41666666666667
/sin(17*x)\ lim |---------| x->0-\sin(12*x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
17 -- 12
$$\frac{17}{12}$$
= 1.41666666666667
= 1.41666666666667
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{17}{12}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{17}{12}$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{\sin{\left(17 \right)}}{\sin{\left(12 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{\sin{\left(17 \right)}}{\sin{\left(12 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{17}{12}$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{\sin{\left(17 \right)}}{\sin{\left(12 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right) = \frac{\sin{\left(17 \right)}}{\sin{\left(12 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(17 x \right)}}{\sin{\left(12 x \right)}}\right)$$
More at x→-oo
The graph