Mister Exam
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How to use it?
- Limit of the function:
-
Limit of ((3+2*x)/(7+5*x))^(1+x)
-
Limit of (4+x^2-5*x)/(-16+x^2)
-
Limit of sin(2*x)/tan(x)
-
Limit of (5-5*x)/(-1+sqrt(x))
-
Identical expressions
- sin(two *x)/tan(x)
- sinus of (2 multiply by x) divide by tangent of (x)
- sinus of (two multiply by x) divide by tangent of (x)
- sin(2x)/tan(x)
- sin2x/tanx
- sin(2*x) divide by tan(x)
-
Similar expressions
- asin(2*x)/tan(x/2)
Limit of the function sin(2*x)/tan(x)
The solution
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/sin(2*x)\ lim |--------| x->0+\ tan(x) /
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
Limit(sin(2*x)/tan(x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x} \frac{x}{\tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(x \right)}}\right)$$
=
Do replacement
$$u = 2 x$$
and
$$v = x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{2 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = 2$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x} \frac{x}{\tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(x \right)}}\right)$$
=
Do replacement
$$u = 2 x$$
and
$$v = x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{2 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = 2$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(2 x \right)}}{\frac{d}{d x} \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \cos{\left(2 x \right)}}{\tan^{2}{\left(x \right)} + 1}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2}{\tan^{2}{\left(x \right)} + 1}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2}{\tan^{2}{\left(x \right)} + 1}\right)$$
=
$$2$$
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
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(2 x \right)}}{\frac{d}{d x} \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \cos{\left(2 x \right)}}{\tan^{2}{\left(x \right)} + 1}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2}{\tan^{2}{\left(x \right)} + 1}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2}{\tan^{2}{\left(x \right)} + 1}\right)$$
=
$$2$$
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(2*x)\ lim |--------| x->0+\ tan(x) /
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
2
$$2$$
= 2.0
/sin(2*x)\ lim |--------| x->0-\ tan(x) /
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
2
$$2$$
= 2.0
= 2.0
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = 2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = 2$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \frac{\sin{\left(2 \right)}}{\tan{\left(1 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \frac{\sin{\left(2 \right)}}{\tan{\left(1 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = 2$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \frac{\sin{\left(2 \right)}}{\tan{\left(1 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right) = \frac{\sin{\left(2 \right)}}{\tan{\left(1 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(2 x \right)}}{\tan{\left(x \right)}}\right)$$
More at x→-oo
The graph