Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of tan(8*x)/(5*x)
-
Limit of x^6
-
Limit of cos(x)^4
-
Limit of ((1+2*x)^4-(-1+x)^4)/((1+2*x)^4+(-1+x)^4)
-
Identical expressions
- tan(eight *x)/(five *x)
- tangent of (8 multiply by x) divide by (5 multiply by x)
- tangent of (eight multiply by x) divide by (five multiply by x)
- tan(8x)/(5x)
- tan8x/5x
- tan(8*x) divide by (5*x)
-
What you mean?
- tan(8*x)/((5*x))
- tan(8*x)*x/5
- tan(8*x)/((5*x))
- tan(8*x)*x/5
- tan(8*x)/((5*x))
- tan(8*x)*x/5
- tan(8*x)*x/5
You entered:
tan(8*x)/(5*x)
What you mean?
Limit of the function tan(8*x)/(5*x)
The solution
You have entered
[src]
/tan(8*x)\ lim |--------| x->0+\ 5*x /
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
Limit(tan(8*x)/((5*x)), x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \lim_{x \to 0^+}\left(\frac{\frac{1}{5 x} \sin{\left(8 x \right)}}{\cos{\left(8 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{1}{5 x} \sin{\left(8 x \right)}\right) \lim_{x \to 0^+} \frac{1}{\cos{\left(8 x \right)}} = \lim_{x \to 0^+}\left(\frac{1}{5 x} \sin{\left(8 x \right)}\right)$$
Do replacement
$$u = 8 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(8 x \right)}}{5 x}\right) = \lim_{u \to 0^+}\left(\frac{8 \sin{\left(u \right)}}{5 u}\right)$$
=
$$\frac{8 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{5}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{8}{5}$$
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
transform
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \lim_{x \to 0^+}\left(\frac{\frac{1}{5 x} \sin{\left(8 x \right)}}{\cos{\left(8 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{1}{5 x} \sin{\left(8 x \right)}\right) \lim_{x \to 0^+} \frac{1}{\cos{\left(8 x \right)}} = \lim_{x \to 0^+}\left(\frac{1}{5 x} \sin{\left(8 x \right)}\right)$$
Do replacement
$$u = 8 x$$
then
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(8 x \right)}}{5 x}\right) = \lim_{u \to 0^+}\left(\frac{8 \sin{\left(u \right)}}{5 u}\right)$$
=
$$\frac{8 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{5}$$
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:
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{8}{5}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \tan{\left(8 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+}\left(5 x\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan{\left(8 x \right)}}{\frac{d}{d x} 5 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{8 \tan^{2}{\left(8 x \right)}}{5} + \frac{8}{5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{8 \tan^{2}{\left(8 x \right)}}{5} + \frac{8}{5}\right)$$
=
$$\frac{8}{5}$$
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
$$\lim_{x \to 0^+} \tan{\left(8 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+}\left(5 x\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan{\left(8 x \right)}}{\frac{d}{d x} 5 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{8 \tan^{2}{\left(8 x \right)}}{5} + \frac{8}{5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{8 \tan^{2}{\left(8 x \right)}}{5} + \frac{8}{5}\right)$$
=
$$\frac{8}{5}$$
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(8*x)\ lim |--------| x->0+\ 5*x /
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
8/5
$$\frac{8}{5}$$
= 1.6
/tan(8*x)\ lim |--------| x->0-\ 5*x /
$$\lim_{x \to 0^-}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
8/5
$$\frac{8}{5}$$
= 1.6
= 1.6
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{8}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{8}{5}$$
$$\lim_{x \to \infty}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{\tan{\left(8 \right)}}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{\tan{\left(8 \right)}}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{8}{5}$$
$$\lim_{x \to \infty}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{\tan{\left(8 \right)}}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right) = \frac{\tan{\left(8 \right)}}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\tan{\left(8 x \right)}}{5 x}\right)$$
More at x→-oo
The graph