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