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sqrt(4+n)-sqrt(-1+n)
Limit of the function/ sqrt(4+n)-sqrt(-1+n)

Limit of the function sqrt(4+n)-sqrt(-1+n)

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     /  _______     ________\
 lim \\/ 4 + n  - \/ -1 + n /
n->oo
lim⁡n→∞(−n−1+n+4)\lim_{n \to \infty}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) 
Limit(sqrt(4 + n) - sqrt(-1 + n), n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞(−n−1+n+4)\lim_{n \to \infty}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right)
Let's eliminate indeterminateness oo - oo
Multiply and divide by
n−1+n+4\sqrt{n - 1} + \sqrt{n + 4}
then
lim⁡n→∞(−n−1+n+4)\lim_{n \to \infty}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right)
=
lim⁡n→∞((−n−1+n+4)(n−1+n+4)n−1+n+4)\lim_{n \to \infty}\left(\frac{\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) \left(\sqrt{n - 1} + \sqrt{n + 4}\right)}{\sqrt{n - 1} + \sqrt{n + 4}}\right)
=
lim⁡n→∞(−(n−1)2+(n+4)2n−1+n+4)\lim_{n \to \infty}\left(\frac{- \left(\sqrt{n - 1}\right)^{2} + \left(\sqrt{n + 4}\right)^{2}}{\sqrt{n - 1} + \sqrt{n + 4}}\right)
=
lim⁡n→∞((1−n)+(n+4)n−1+n+4)\lim_{n \to \infty}\left(\frac{\left(1 - n\right) + \left(n + 4\right)}{\sqrt{n - 1} + \sqrt{n + 4}}\right)
=
lim⁡n→∞(5n−1+n+4)\lim_{n \to \infty}\left(\frac{5}{\sqrt{n - 1} + \sqrt{n + 4}}\right)

Let's divide numerator and denominator by sqrt(n):
lim⁡n→∞(5n(n−1n+n+4n))\lim_{n \to \infty}\left(\frac{5}{\sqrt{n} \left(\frac{\sqrt{n - 1}}{\sqrt{n}} + \frac{\sqrt{n + 4}}{\sqrt{n}}\right)}\right) =
lim⁡n→∞(5n(n−1n+n+4n))\lim_{n \to \infty}\left(\frac{5}{\sqrt{n} \left(\sqrt{\frac{n - 1}{n}} + \sqrt{\frac{n + 4}{n}}\right)}\right) =
lim⁡n→∞(5n(1−1n+1+4n))\lim_{n \to \infty}\left(\frac{5}{\sqrt{n} \left(\sqrt{1 - \frac{1}{n}} + \sqrt{1 + \frac{4}{n}}\right)}\right)
Do replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(5n(1−1n+1+4n))\lim_{n \to \infty}\left(\frac{5}{\sqrt{n} \left(\sqrt{1 - \frac{1}{n}} + \sqrt{1 + \frac{4}{n}}\right)}\right) =
lim⁡u→0+(5(1−u+4u+1)1u)\lim_{u \to 0^+}\left(\frac{5}{\left(\sqrt{1 - u} + \sqrt{4 u + 1}\right) \sqrt{\frac{1}{u}}}\right) =
= 5∞~(1−0+0⋅4+1)=0\frac{5}{\tilde{\infty} \left(\sqrt{1 - 0} + \sqrt{0 \cdot 4 + 1}\right)} = 0

The final answer:
lim⁡n→∞(−n−1+n+4)=0\lim_{n \to \infty}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) = 0
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
02468-8-6-4-2-101002
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Rapid solution [src] 
0
00
Expand and simplify
Other limits n→0, -oo, +oo, 1
lim⁡n→∞(−n−1+n+4)=0\lim_{n \to \infty}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) = 0
lim⁡n→0−(−n−1+n+4)=2−i\lim_{n \to 0^-}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) = 2 - i
More at n→0 from the left
lim⁡n→0+(−n−1+n+4)=2−i\lim_{n \to 0^+}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) = 2 - i
More at n→0 from the right
lim⁡n→1−(−n−1+n+4)=5\lim_{n \to 1^-}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) = \sqrt{5}
More at n→1 from the left
lim⁡n→1+(−n−1+n+4)=5\lim_{n \to 1^+}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) = \sqrt{5}
More at n→1 from the right
lim⁡n→−∞(−n−1+n+4)=0\lim_{n \to -\infty}\left(- \sqrt{n - 1} + \sqrt{n + 4}\right) = 0
More at n→-oo
The graph
Limit of the function sqrt(4+n)-sqrt(-1+n)
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