Mister Exam
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How to use it?
- Limit of the function:
-
Limit of x^2*(1/3-cos(8*x)/3)
-
Limit of (-2+x^2-x)/(-2+x+3*x^2)
-
Limit of (-1+sin(x))/cos(x)
-
Limit of (-2*sin(x)+sin(2*x))/(x*log(cos(5*x)))
- Derivative of:
- sin(10*x)/tan(5*x)
-
Identical expressions
- sin(ten *x)/tan(five *x)
- sinus of (10 multiply by x) divide by tangent of (5 multiply by x)
- sinus of (ten multiply by x) divide by tangent of (five multiply by x)
- sin(10x)/tan(5x)
- sin10x/tan5x
- sin(10*x) divide by tan(5*x)
Limit of the function sin(10*x)/tan(5*x)
The solution
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[src]
/sin(10*x)\ lim |---------| x->0+\ tan(5*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
Limit(sin(10*x)/tan(5*x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{x} \frac{x}{\tan{\left(5 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(5 x \right)}}\right)$$
=
Do replacement
$$u = 10 x$$
and
$$v = 5 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{10 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{5 \tan{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = 2$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{x} \frac{x}{\tan{\left(5 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(5 x \right)}}\right)$$
=
Do replacement
$$u = 10 x$$
and
$$v = 5 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{10 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{5 \tan{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)$$
=
$$2 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = 2$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(10 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(5 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(10 x \right)}}{\frac{d}{d x} \tan{\left(5 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{10 \cos{\left(10 x \right)}}{5 \tan^{2}{\left(5 x \right)} + 5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{10}{5 \tan^{2}{\left(5 x \right)} + 5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{10}{5 \tan^{2}{\left(5 x \right)} + 5}\right)$$
=
$$2$$
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
$$\lim_{x \to 0^+} \sin{\left(10 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(5 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(10 x \right)}}{\frac{d}{d x} \tan{\left(5 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{10 \cos{\left(10 x \right)}}{5 \tan^{2}{\left(5 x \right)} + 5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{10}{5 \tan^{2}{\left(5 x \right)} + 5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{10}{5 \tan^{2}{\left(5 x \right)} + 5}\right)$$
=
$$2$$
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
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = 2$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = 2$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \frac{\sin{\left(10 \right)}}{\tan{\left(5 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \frac{\sin{\left(10 \right)}}{\tan{\left(5 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = 2$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \frac{\sin{\left(10 \right)}}{\tan{\left(5 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right) = \frac{\sin{\left(10 \right)}}{\tan{\left(5 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
More at x→-oo
One‐sided limits
[src]
/sin(10*x)\ lim |---------| x->0+\ tan(5*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
2
$$2$$
= 2.0
/sin(10*x)\ lim |---------| x->0-\ tan(5*x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(10 x \right)}}{\tan{\left(5 x \right)}}\right)$$
2
$$2$$
= 2.0
= 2.0
The graph