Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of tan(5*x)/sin(2*x)
-
Limit of -x
-
Limit of x/sqrt(1+x^2)
- Limit of x+y
-
Identical expressions
- tan(five *x)/sin(two *x)
- tangent of (5 multiply by x) divide by sinus of (2 multiply by x)
- tangent of (five multiply by x) divide by sinus of (two multiply by x)
- tan(5x)/sin(2x)
- tan5x/sin2x
- tan(5*x) divide by sin(2*x)
Limit of the function tan(5*x)/sin(2*x)
The solution
You have entered
[src]
/tan(5*x)\ lim |--------| x->0+\sin(2*x)/
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(5 x \right)}}{\sin{\left(2 x \right)}}\right)$$
Limit(tan(5*x)/sin(2*x), x, 0)
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \tan{\left(5 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(5 x \right)}}{\sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan{\left(5 x \right)}}{\frac{d}{d x} \sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5 \tan^{2}{\left(5 x \right)} + 5}{2 \cos{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5 \tan^{2}{\left(5 x \right)}}{2} + \frac{5}{2}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5 \tan^{2}{\left(5 x \right)}}{2} + \frac{5}{2}\right)$$
=
$$\frac{5}{2}$$
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
$$\lim_{x \to 0^+} \tan{\left(5 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(5 x \right)}}{\sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan{\left(5 x \right)}}{\frac{d}{d x} \sin{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5 \tan^{2}{\left(5 x \right)} + 5}{2 \cos{\left(2 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5 \tan^{2}{\left(5 x \right)}}{2} + \frac{5}{2}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5 \tan^{2}{\left(5 x \right)}}{2} + \frac{5}{2}\right)$$
=
$$\frac{5}{2}$$
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]
/tan(5*x)\ lim |--------| x->0+\sin(2*x)/
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(5 x \right)}}{\sin{\left(2 x \right)}}\right)$$
5/2
$$\frac{5}{2}$$
= 2.5
/tan(5*x)\ lim |--------| x->0-\sin(2*x)/
$$\lim_{x \to 0^-}\left(\frac{\tan{\left(5 x \right)}}{\sin{\left(2 x \right)}}\right)$$
5/2
$$\frac{5}{2}$$
= 2.5
= 2.5
The graph