Mister Exam
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How to use it?
- Limit of the function:
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Limit of (-3+sqrt(1+4*x))/(-2+x)
-
Limit of 1/(-8+x)
-
Limit of (1/x)^(log(1+x)/log(2-x))
-
Limit of (1-cos(6*x))/(1-cos(8*x))
-
Identical expressions
- sin(three *x)/tan(three *x)
- sinus of (3 multiply by x) divide by tangent of (3 multiply by x)
- sinus of (three multiply by x) divide by tangent of (three multiply by x)
- sin(3x)/tan(3x)
- sin3x/tan3x
- sin(3*x) divide by tan(3*x)
Limit of the function sin(3*x)/tan(3*x)
The solution
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[src]
/sin(3*x)\ lim |--------| x->0+\tan(3*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
Limit(sin(3*x)/tan(3*x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x} \frac{x}{\tan{\left(3 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(3 x \right)}}\right)$$
=
Do replacement
$$u = 3 x$$
and
$$v = 3 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{3 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{3 \tan{\left(v \right)}}\right)$$
=
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)$$
=
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}$$
transform
$$\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right) = \lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v \cos{\left(v \right)}}\right)$$
=
$$\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right) \lim_{v \to 0^+} \frac{1}{\cos{\left(v \right)}} = \lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
is first remarkable limit, is equal to 1.
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = 1$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x} \frac{x}{\tan{\left(3 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(3 x \right)}}\right)$$
=
Do replacement
$$u = 3 x$$
and
$$v = 3 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{3 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{3 \tan{\left(v \right)}}\right)$$
=
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)$$
=
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}$$
transform
$$\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right) = \lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v \cos{\left(v \right)}}\right)$$
=
$$\lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right) \lim_{v \to 0^+} \frac{1}{\cos{\left(v \right)}} = \lim_{v \to 0^+}\left(\frac{\sin{\left(v \right)}}{v}\right)$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
is first remarkable limit, is equal to 1.
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = 1$$
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
$$\lim_{x \to 0^+} \tan{\left(3 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(3 x \right)}}{\frac{d}{d x} \tan{\left(3 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3 \cos{\left(3 x \right)}}{3 \tan^{2}{\left(3 x \right)} + 3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3}{3 \tan^{2}{\left(3 x \right)} + 3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3}{3 \tan^{2}{\left(3 x \right)} + 3}\right)$$
=
$$1$$
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
$$\lim_{x \to 0^+} \tan{\left(3 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(3 x \right)}}{\frac{d}{d x} \tan{\left(3 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3 \cos{\left(3 x \right)}}{3 \tan^{2}{\left(3 x \right)} + 3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3}{3 \tan^{2}{\left(3 x \right)} + 3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3}{3 \tan^{2}{\left(3 x \right)} + 3}\right)$$
=
$$1$$
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+\tan(3*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
1
$$1$$
= 1.0
/sin(3*x)\ lim |--------| x->0-\tan(3*x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
1
$$1$$
= 1.0
= 1.0
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = 1$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = \frac{\sin{\left(3 \right)}}{\tan{\left(3 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = \frac{\sin{\left(3 \right)}}{\tan{\left(3 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = 1$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = \frac{\sin{\left(3 \right)}}{\tan{\left(3 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right) = \frac{\sin{\left(3 \right)}}{\tan{\left(3 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(3 x \right)}}{\tan{\left(3 x \right)}}\right)$$
More at x→-oo
The graph