Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of 7^(1/(-3+x))
-
Limit of (3-3*x^2+4*x^4+6*x^3)/(2*x^2+7*x^4)
-
Limit of ((5+4*x)/(-1+5*x))^(1+3*x)
-
Limit of (-6-x^2-3*x+4*x^3)/(3-x^2+2*x^3)
-
Identical expressions
- sqrt(one +n)/sqrt(two +n)
- square root of (1 plus n) divide by square root of (2 plus n)
- square root of (one plus n) divide by square root of (two plus n)
- √(1+n)/√(2+n)
- sqrt1+n/sqrt2+n
- sqrt(1+n) divide by sqrt(2+n)
-
Similar expressions
- sqrt(1-n)/sqrt(2+n)
- sqrt(1+n)/sqrt(2-n)
Limit of the function sqrt(1+n)/sqrt(2+n)
The solution
You have entered
[src]
/ _______\
|\/ 1 + n |
lim |---------|
n->oo| _______|
\\/ 2 + n /
$$\lim_{n \to \infty}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right)$$
Limit(sqrt(1 + n)/sqrt(2 + n), n, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty} \sqrt{n + 1} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} \sqrt{n + 2} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \sqrt{n + 1}}{\frac{d}{d n} \sqrt{n + 2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\sqrt{n + 2}}{\sqrt{n + 1}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\sqrt{n + 2}}{\sqrt{n + 1}}\right)$$
=
$$1$$
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
$$\lim_{n \to \infty} \sqrt{n + 1} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} \sqrt{n + 2} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \sqrt{n + 1}}{\frac{d}{d n} \sqrt{n + 2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\sqrt{n + 2}}{\sqrt{n + 1}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\sqrt{n + 2}}{\sqrt{n + 1}}\right)$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{2}}{2}$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{2}}{2}$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{6}}{3}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{6}}{3}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{2}}{2}$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{2}}{2}$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{6}}{3}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{6}}{3}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = 1$$
More at n→-oo
The graph