Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of ((1+3*x)/(2+3*x))^(5*x)
-
Limit of (x*(1/2+x))^x
-
Limit of (2-sqrt(x))/(4-sqrt(2)*sqrt(x))
-
Limit of (5+2*n)/n
-
Identical expressions
- (five + two *n)/n
- (5 plus 2 multiply by n) divide by n
- (five plus two multiply by n) divide by n
- (5+2n)/n
- 5+2n/n
- (5+2*n) divide by n
-
Similar expressions
- (5-2*n)/n
- (-5+2*n)/n
Limit of the function (5+2*n)/n
The solution
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/5 + 2*n\ lim |-------| n->oo\ n /
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right)$$
Limit((5 + 2*n)/n, n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{2 + \frac{5}{n}}{1}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{2 + \frac{5}{n}}{1}\right) = \lim_{u \to 0^+}\left(5 u + 2\right)$$
=
$$0 \cdot 5 + 2 = 2$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right) = 2$$
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{2 + \frac{5}{n}}{1}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{2 + \frac{5}{n}}{1}\right) = \lim_{u \to 0^+}\left(5 u + 2\right)$$
=
$$0 \cdot 5 + 2 = 2$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right) = 2$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(2 n + 5\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} n = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(2 n + 5\right)}{\frac{d}{d n} n}\right)$$
=
$$\lim_{n \to \infty} 2$$
=
$$\lim_{n \to \infty} 2$$
=
$$2$$
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
$$\lim_{n \to \infty}\left(2 n + 5\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} n = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(2 n + 5\right)}{\frac{d}{d n} n}\right)$$
=
$$\lim_{n \to \infty} 2$$
=
$$\lim_{n \to \infty} 2$$
=
$$2$$
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
$$\lim_{n \to \infty}\left(\frac{2 n + 5}{n}\right) = 2$$
$$\lim_{n \to 0^-}\left(\frac{2 n + 5}{n}\right) = -\infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{2 n + 5}{n}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{2 n + 5}{n}\right) = 7$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{2 n + 5}{n}\right) = 7$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{2 n + 5}{n}\right) = 2$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{2 n + 5}{n}\right) = -\infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{2 n + 5}{n}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{2 n + 5}{n}\right) = 7$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{2 n + 5}{n}\right) = 7$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{2 n + 5}{n}\right) = 2$$
More at n→-oo
The graph