Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of (cos(x)+sin(x))^(1/x)
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Limit of (2/3)^x
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Limit of (3^n+4^n)/(3^(1+n)+4^(1+n))
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Limit of (1+n)*(4+n)/(n*(3+n))
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Identical expressions
- (one +n)*(four +n)/(n*(three +n))
- (1 plus n) multiply by (4 plus n) divide by (n multiply by (3 plus n))
- (one plus n) multiply by (four plus n) divide by (n multiply by (three plus n))
- (1+n)(4+n)/(n(3+n))
- 1+n4+n/n3+n
- (1+n)*(4+n) divide by (n*(3+n))
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Similar expressions
- (1+n)*(4-n)/(n*(3+n))
- (1-n)*(4+n)/(n*(3+n))
- (1+n)*(4+n)/(n*(3-n))
Limit of the function (1+n)*(4+n)/(n*(3+n))
The solution
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/(1 + n)*(4 + n)\ lim |---------------| n->oo\ n*(3 + n) /
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right)$$
Limit(((1 + n)*(4 + n))/((n*(3 + n))), n, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n}\right) = \infty$$
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.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{\left(n + 1\right) \left(n + 4\right)}{n}}{\frac{d}{d n} \left(n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(1 - \frac{4}{n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(1 - \frac{4}{n^{2}}\right)$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n}\right) = \infty$$
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.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{\left(n + 1\right) \left(n + 4\right)}{n}}{\frac{d}{d n} \left(n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(1 - \frac{4}{n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(1 - \frac{4}{n^{2}}\right)$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = -\infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = \frac{5}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = \frac{5}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = -\infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = \frac{5}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = \frac{5}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 3\right)}\right) = 1$$
More at n→-oo
The graph