Mister Exam
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How to use it?
- Limit of the function:
-
Limit of ((-2+x)/x)^(2*x)
-
Limit of (-1+tan(x)^(1/3))/(-1+log(x)/log(pi/4))
-
Limit of (-5+sqrt(9+x^2))/(-3+sqrt(5+x))
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Limit of sin(7*x)/tan(4*x)
-
Identical expressions
- sin(seven *x)/tan(four *x)
- sinus of (7 multiply by x) divide by tangent of (4 multiply by x)
- sinus of (seven multiply by x) divide by tangent of (four multiply by x)
- sin(7x)/tan(4x)
- sin7x/tan4x
- sin(7*x) divide by tan(4*x)
Limit of the function sin(7*x)/tan(4*x)
The solution
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/sin(7*x)\ lim |--------| x->0+\tan(4*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
Limit(sin(7*x)/tan(4*x), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{x} \frac{x}{\tan{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(4 x \right)}}\right)$$
=
Do replacement
$$u = 7 x$$
and
$$v = 4 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{7 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{4 \tan{\left(v \right)}}\right)$$
=
$$\frac{7 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)}{4}$$
=
$$\frac{7 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}}{4}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{7}{4}$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{x} \frac{x}{\tan{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{x}\right) \lim_{x \to 0^+}\left(\frac{x}{\tan{\left(4 x \right)}}\right)$$
=
Do replacement
$$u = 7 x$$
and
$$v = 4 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \lim_{u \to 0^+}\left(\frac{7 \sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{4 \tan{\left(v \right)}}\right)$$
=
$$\frac{7 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{v \to 0^+}\left(\frac{v}{\tan{\left(v \right)}}\right)}{4}$$
=
$$\frac{7 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \left(\lim_{v \to 0^+}\left(\frac{\tan{\left(v \right)}}{v}\right)\right)^{-1}}{4}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{7}{4}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(7 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(4 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(7 x \right)}}{\frac{d}{d x} \tan{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{7 \cos{\left(7 x \right)}}{4 \tan^{2}{\left(4 x \right)} + 4}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{7}{4 \tan^{2}{\left(4 x \right)} + 4}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{7}{4 \tan^{2}{\left(4 x \right)} + 4}\right)$$
=
$$\frac{7}{4}$$
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
$$\lim_{x \to 0^+} \sin{\left(7 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(4 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(7 x \right)}}{\frac{d}{d x} \tan{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{7 \cos{\left(7 x \right)}}{4 \tan^{2}{\left(4 x \right)} + 4}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{7}{4 \tan^{2}{\left(4 x \right)} + 4}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{7}{4 \tan^{2}{\left(4 x \right)} + 4}\right)$$
=
$$\frac{7}{4}$$
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(7*x)\ lim |--------| x->0+\tan(4*x)/
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
7/4
$$\frac{7}{4}$$
= 1.75
/sin(7*x)\ lim |--------| x->0-\tan(4*x)/
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
7/4
$$\frac{7}{4}$$
= 1.75
= 1.75
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{7}{4}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{7}{4}$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{\sin{\left(7 \right)}}{\tan{\left(4 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{\sin{\left(7 \right)}}{\tan{\left(4 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{7}{4}$$
$$\lim_{x \to \infty}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{\sin{\left(7 \right)}}{\tan{\left(4 \right)}}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right) = \frac{\sin{\left(7 \right)}}{\tan{\left(4 \right)}}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(7 x \right)}}{\tan{\left(4 x \right)}}\right)$$
More at x→-oo
The graph