Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-1+3*sqrt(x))/(-1+4*sqrt(x))
-
Limit of (-4+2*x)/(6+4*x+7*x^2)
-
Limit of ((5+2*x)/(2*x))^(3*x)
-
Limit of ((-4+2*x)/(1+2*x))^(5+3*x)
-
Identical expressions
- (two +n)^ two /(n*(three +n))
- (2 plus n) squared divide by (n multiply by (3 plus n))
- (two plus n) to the power of two divide by (n multiply by (three plus n))
- (2+n)2/(n*(3+n))
- 2+n2/n*3+n
- (2+n)²/(n*(3+n))
- (2+n) to the power of 2/(n*(3+n))
- (2+n)^2/(n(3+n))
- (2+n)2/(n(3+n))
- 2+n2/n3+n
- 2+n^2/n3+n
- (2+n)^2 divide by (n*(3+n))
-
Similar expressions
- (2+n)^2/(n*(3-n))
- (2-n)^2/(n*(3+n))
Limit of the function (2+n)^2/(n*(3+n))
The solution
You have entered
[src]
/ 2\
| (2 + n) |
lim |---------|
n->oo\n*(3 + n)/
$$\lim_{n \to \infty}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right)$$
Limit((2 + n)^2/((n*(3 + n))), n, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(\frac{\left(n + 2\right)^{2}}{n}\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(n + 3\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{\left(n + 2\right)^{2}}{n}}{\frac{d}{d n} \left(n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(1 - \frac{4}{n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(1 - \frac{4}{n^{2}}\right)$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(\frac{\left(n + 2\right)^{2}}{n}\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(n + 3\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{\left(n + 2\right)^{2}}{n}}{\frac{d}{d n} \left(n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(1 - \frac{4}{n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(1 - \frac{4}{n^{2}}\right)$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = -\infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = \frac{9}{4}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = \frac{9}{4}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = -\infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = \frac{9}{4}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = \frac{9}{4}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\left(n + 2\right)^{2}}{n \left(n + 3\right)}\right) = 1$$
More at n→-oo
The graph