Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of cos(x^2)
-
Limit of (-sin(x)+cos(x))^(1/x)
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Limit of (cos(x)/cosh(x))^(1/(x*log(1+x)))
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Limit of (-1+cos(x))*cot(x)
- Graphing y =:
- sqrt(n+n^2)-sqrt(-1+n^2)
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Identical expressions
- sqrt(n+n^ two)-sqrt(- one +n^ two)
- square root of (n plus n squared ) minus square root of ( minus 1 plus n squared )
- square root of (n plus n to the power of two) minus square root of ( minus one plus n to the power of two)
- √(n+n^2)-√(-1+n^2)
- sqrt(n+n2)-sqrt(-1+n2)
- sqrtn+n2-sqrt-1+n2
- sqrt(n+n²)-sqrt(-1+n²)
- sqrt(n+n to the power of 2)-sqrt(-1+n to the power of 2)
- sqrtn+n^2-sqrt-1+n^2
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Similar expressions
- sqrt(n-n^2)-sqrt(-1+n^2)
- sqrt(n+n^2)-sqrt(1+n^2)
- sqrt(n+n^2)+sqrt(-1+n^2)
- sqrt(n+n^2)-sqrt(-1-n^2)
Limit of the function sqrt(n+n^2)-sqrt(-1+n^2)
The solution
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/ ________ _________\
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lim \\/ n + n - \/ -1 + n /
n->oo
$$\lim_{n \to \infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right)$$
Limit(sqrt(n + n^2) - sqrt(-1 + n^2), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right)$$
Let's eliminate indeterminateness oo - oo
Multiply and divide by
$$\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}$$
then
$$\lim_{n \to \infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) \left(\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right)}{\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{- \left(\sqrt{n^{2} - 1}\right)^{2} + \left(\sqrt{n^{2} + n}\right)^{2}}{\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(1 - n^{2}\right) + \left(n^{2} + n\right)}{\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n + 1}{\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\frac{\sqrt{n^{2} - 1}}{n} + \frac{\sqrt{n^{2} + n}}{n}}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\sqrt{\frac{n^{2} - 1}{n^{2}}} + \sqrt{\frac{n^{2} + n}{n^{2}}}}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\sqrt{1 - \frac{1}{n^{2}}} + \sqrt{1 + \frac{1}{n}}}\right)$$
Do replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\sqrt{1 - \frac{1}{n^{2}}} + \sqrt{1 + \frac{1}{n}}}\right)$$ =
$$\lim_{u \to 0^+}\left(\frac{u + 1}{\sqrt{1 - u^{2}} + \sqrt{u + 1}}\right)$$ =
= $$\frac{1}{\sqrt{1} + \sqrt{1 - 0^{2}}} = \frac{1}{2}$$
The final answer:
$$\lim_{n \to \infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = \frac{1}{2}$$
$$\lim_{n \to \infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right)$$
Let's eliminate indeterminateness oo - oo
Multiply and divide by
$$\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}$$
then
$$\lim_{n \to \infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) \left(\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right)}{\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{- \left(\sqrt{n^{2} - 1}\right)^{2} + \left(\sqrt{n^{2} + n}\right)^{2}}{\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(1 - n^{2}\right) + \left(n^{2} + n\right)}{\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n + 1}{\sqrt{n^{2} - 1} + \sqrt{n^{2} + n}}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\frac{\sqrt{n^{2} - 1}}{n} + \frac{\sqrt{n^{2} + n}}{n}}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\sqrt{\frac{n^{2} - 1}{n^{2}}} + \sqrt{\frac{n^{2} + n}{n^{2}}}}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\sqrt{1 - \frac{1}{n^{2}}} + \sqrt{1 + \frac{1}{n}}}\right)$$
Do replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\sqrt{1 - \frac{1}{n^{2}}} + \sqrt{1 + \frac{1}{n}}}\right)$$ =
$$\lim_{u \to 0^+}\left(\frac{u + 1}{\sqrt{1 - u^{2}} + \sqrt{u + 1}}\right)$$ =
= $$\frac{1}{\sqrt{1} + \sqrt{1 - 0^{2}}} = \frac{1}{2}$$
The final answer:
$$\lim_{n \to \infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = \frac{1}{2}$$
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
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = \frac{1}{2}$$
$$\lim_{n \to 0^-}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = - i$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = - i$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = \sqrt{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = \sqrt{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = - \frac{1}{2}$$
More at n→-oo
$$\lim_{n \to 0^-}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = - i$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = - i$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = \sqrt{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = \sqrt{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(- \sqrt{n^{2} - 1} + \sqrt{n^{2} + n}\right) = - \frac{1}{2}$$
More at n→-oo
The graph