Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of ((-5+2*x)/(3+2*x))^(7*x)
-
Limit of (1-3*x+2*x^2)/(4+x+3*x^2)
-
Limit of (-28+x^2+3*x)/(-64+x^3)
-
Limit of (3+5*n)/(1+n)
-
Identical expressions
- (three + five *n)/(one +n)
- (3 plus 5 multiply by n) divide by (1 plus n)
- (three plus five multiply by n) divide by (one plus n)
- (3+5n)/(1+n)
- 3+5n/1+n
- (3+5*n) divide by (1+n)
-
Similar expressions
- (3-5*n)/(1+n)
- (3+5*n)/(1-n)
Limit of the function (3+5*n)/(1+n)
The solution
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/3 + 5*n\ lim |-------| n->oo\ 1 + n /
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right)$$
Limit((3 + 5*n)/(1 + n), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{5 + \frac{3}{n}}{1 + \frac{1}{n}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{5 + \frac{3}{n}}{1 + \frac{1}{n}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 5}{u + 1}\right)$$
=
$$\frac{0 \cdot 3 + 5}{1} = 5$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right) = 5$$
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{5 + \frac{3}{n}}{1 + \frac{1}{n}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{5 + \frac{3}{n}}{1 + \frac{1}{n}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 5}{u + 1}\right)$$
=
$$\frac{0 \cdot 3 + 5}{1} = 5$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right) = 5$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(5 n + 3\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(n + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(5 n + 3\right)}{\frac{d}{d n} \left(n + 1\right)}\right)$$
=
$$\lim_{n \to \infty} 5$$
=
$$\lim_{n \to \infty} 5$$
=
$$5$$
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
$$\lim_{n \to \infty}\left(5 n + 3\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(n + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(5 n + 3\right)}{\frac{d}{d n} \left(n + 1\right)}\right)$$
=
$$\lim_{n \to \infty} 5$$
=
$$\lim_{n \to \infty} 5$$
=
$$5$$
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
$$\lim_{n \to \infty}\left(\frac{5 n + 3}{n + 1}\right) = 5$$
$$\lim_{n \to 0^-}\left(\frac{5 n + 3}{n + 1}\right) = 3$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{5 n + 3}{n + 1}\right) = 3$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{5 n + 3}{n + 1}\right) = 4$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{5 n + 3}{n + 1}\right) = 4$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{5 n + 3}{n + 1}\right) = 5$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{5 n + 3}{n + 1}\right) = 3$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{5 n + 3}{n + 1}\right) = 3$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{5 n + 3}{n + 1}\right) = 4$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{5 n + 3}{n + 1}\right) = 4$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{5 n + 3}{n + 1}\right) = 5$$
More at n→-oo
The graph