Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
- Limit of (9+x^2-6*x)/(6+x^2-5*x)
-
Limit of (9+x^2+6*x)/(-3+x^2+2*x)
-
Limit of (-64+x^2)/(8+x)
-
Limit of 4*x^2+11*x
- Sum of series:
-
2*n/(1+n^2)
-
Identical expressions
- two *n/(one +n^ two)
- 2 multiply by n divide by (1 plus n squared )
- two multiply by n divide by (one plus n to the power of two)
- 2*n/(1+n2)
- 2*n/1+n2
- 2*n/(1+n²)
- 2*n/(1+n to the power of 2)
- 2n/(1+n^2)
- 2n/(1+n2)
- 2n/1+n2
- 2n/1+n^2
- 2*n divide by (1+n^2)
-
Similar expressions
- 2*n/(1-n^2)
Limit of the function 2*n/(1+n^2)
The solution
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[src]
/ 2*n \
lim |------|
n->oo| 2|
\1 + n /
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$
Limit((2*n)/(1 + n^2), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$
Let's divide numerator and denominator by n^2:
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{2 \frac{1}{n}}{1 + \frac{1}{n^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{2 \frac{1}{n}}{1 + \frac{1}{n^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{2 u}{u^{2} + 1}\right)$$
=
$$\frac{0 \cdot 2}{0^{2} + 1} = 0$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$
Let's divide numerator and denominator by n^2:
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{2 \frac{1}{n}}{1 + \frac{1}{n^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{2 \frac{1}{n}}{1 + \frac{1}{n^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{2 u}{u^{2} + 1}\right)$$
=
$$\frac{0 \cdot 2}{0^{2} + 1} = 0$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(2 n\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(n^{2} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 2 n}{\frac{d}{d n} \left(n^{2} + 1\right)}\right)$$
=
$$\lim_{n \to \infty} \frac{1}{n}$$
=
$$\lim_{n \to \infty} \frac{1}{n}$$
=
$$0$$
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)
oo/oo,
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(2 n\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(n^{2} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 2 n}{\frac{d}{d n} \left(n^{2} + 1\right)}\right)$$
=
$$\lim_{n \to \infty} \frac{1}{n}$$
=
$$\lim_{n \to \infty} \frac{1}{n}$$
=
$$0$$
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 n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
$$\lim_{n \to 0^-}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{2 n}{n^{2} + 1}\right) = 1$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{2 n}{n^{2} + 1}\right) = 1$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{2 n}{n^{2} + 1}\right) = 1$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{2 n}{n^{2} + 1}\right) = 1$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{2 n}{n^{2} + 1}\right) = 0$$
More at n→-oo
The graph