Mister Exam
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- Limit of the function:
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Limit of ((1+x)^3+(2+x)^3)/((4+x)^3+(5+x)^3)
-
Limit of 1+(1+x)^2/x^2
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Limit of (sqrt(1-x)-sqrt(1+x))^2/x^2
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Limit of sin(4*x)/tan(7*x)
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Identical expressions
- sin(four *x)/tan(seven *x)
- sinus of (4 multiply by x) divide by tangent of (7 multiply by x)
- sinus of (four multiply by x) divide by tangent of (seven multiply by x)
- sin(4x)/tan(7x)
- sin4x/tan7x
- sin(4*x) divide by tan(7*x)
Limit of the function sin(4*x)/tan(7*x)
The solution
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/sin(4*x)\ lim |--------| x->0+\tan(7*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right)$$
Limit(sin(4*x)/tan(7*x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x} \frac{x}{\tan{\left(7 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(7 x \right)}}\right)$$
=
Do replacement
$$u = 4 x$$
and
$$v = 7 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{4 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{7 \tan{\left(v \right)}}\right)$$
=
$$\frac{4 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)}{7}$$
=
$$\frac{4 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}}{7}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right) = \frac{4}{7}$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x} \frac{x}{\tan{\left(7 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(7 x \right)}}\right)$$
=
Do replacement
$$u = 4 x$$
and
$$v = 7 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{4 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{7 \tan{\left(v \right)}}\right)$$
=
$$\frac{4 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)}{7}$$
=
$$\frac{4 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}}{7}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right) = \frac{4}{7}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(4 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(7 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(4 x \right)}}{\frac{d}{d x} \tan{\left(7 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4 \cos{\left(4 x \right)}}{7 \tan^{2}{\left(7 x \right)} + 7}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4}{7 \tan^{2}{\left(7 x \right)} + 7}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4}{7 \tan^{2}{\left(7 x \right)} + 7}\right)$$
=
$$\frac{4}{7}$$
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
$$\lim_{x \to 0^+} \sin{\left(4 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(7 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(4 x \right)}}{\frac{d}{d x} \tan{\left(7 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4 \cos{\left(4 x \right)}}{7 \tan^{2}{\left(7 x \right)} + 7}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4}{7 \tan^{2}{\left(7 x \right)} + 7}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4}{7 \tan^{2}{\left(7 x \right)} + 7}\right)$$
=
$$\frac{4}{7}$$
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(4*x)\ lim |--------| x->0+\tan(7*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right)$$
4/7
$$\frac{4}{7}$$
= 0.571428571428571
/sin(4*x)\ lim |--------| x->0-\tan(7*x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(4 x \right)}}{\tan{\left(7 x \right)}}\right)$$
4/7
$$\frac{4}{7}$$
= 0.571428571428571
= 0.571428571428571
The graph