Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (9+3*x^2+4*x)/(7-7*x+3*x^2)
-
Limit of (1+2*n)/(-1+3*n)
-
Limit of (x^3-3^x)/(-3+x)
-
Limit of -x+(-2+x)^4/(3+x)^4
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Identical expressions
- (one + two *n)/(- one + three *n)
- (1 plus 2 multiply by n) divide by ( minus 1 plus 3 multiply by n)
- (one plus two multiply by n) divide by ( minus one plus three multiply by n)
- (1+2n)/(-1+3n)
- 1+2n/-1+3n
- (1+2*n) divide by (-1+3*n)
-
Similar expressions
- (1+2*n)/(1+3*n)
- (1-2*n)/(-1+3*n)
- (1+2^n)/(-1+3^n)
- (1+2*n)/(-1-3*n)
Limit of the function (1+2*n)/(-1+3*n)
The solution
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/1 + 2*n \ lim |--------| n->oo\-1 + 3*n/
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right)$$
Limit((1 + 2*n)/(-1 + 3*n), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{2 + \frac{1}{n}}{3 - \frac{1}{n}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{2 + \frac{1}{n}}{3 - \frac{1}{n}}\right) = \lim_{u \to 0^+}\left(\frac{u + 2}{3 - u}\right)$$
=
$$\frac{2}{3 - 0} = \frac{2}{3}$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{2}{3}$$
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{2 + \frac{1}{n}}{3 - \frac{1}{n}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{2 + \frac{1}{n}}{3 - \frac{1}{n}}\right) = \lim_{u \to 0^+}\left(\frac{u + 2}{3 - u}\right)$$
=
$$\frac{2}{3 - 0} = \frac{2}{3}$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{2}{3}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(2 n + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(3 n - 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(2 n + 1\right)}{\frac{d}{d n} \left(3 n - 1\right)}\right)$$
=
$$\lim_{n \to \infty} \frac{2}{3}$$
=
$$\lim_{n \to \infty} \frac{2}{3}$$
=
$$\frac{2}{3}$$
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
$$\lim_{n \to \infty}\left(2 n + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(3 n - 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(2 n + 1\right)}{\frac{d}{d n} \left(3 n - 1\right)}\right)$$
=
$$\lim_{n \to \infty} \frac{2}{3}$$
=
$$\lim_{n \to \infty} \frac{2}{3}$$
=
$$\frac{2}{3}$$
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
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{2}{3}$$
$$\lim_{n \to 0^-}\left(\frac{2 n + 1}{3 n - 1}\right) = -1$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{2 n + 1}{3 n - 1}\right) = -1$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{3}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{3}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{2}{3}$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{2 n + 1}{3 n - 1}\right) = -1$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{2 n + 1}{3 n - 1}\right) = -1$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{3}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{3}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{2 n + 1}{3 n - 1}\right) = \frac{2}{3}$$
More at n→-oo
The graph