Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-2*x^2+4*x^3+5*x)/(3*x^2+7*x)
-
Limit of (-cos(5*x)+cos(3*x))/x^2
-
Limit of (-1+cos(7*x))/(-1+cos(3*x))
-
Limit of (16+x^2+10*x)/(-6+x^2-x)
-
Identical expressions
- (one + two *n)/(three + two *n)
- (1 plus 2 multiply by n) divide by (3 plus 2 multiply by n)
- (one plus two multiply by n) divide by (three plus two multiply by n)
- (1+2n)/(3+2n)
- 1+2n/3+2n
- (1+2*n) divide by (3+2*n)
-
Similar expressions
- ((-1+2*n)/(3+2*n))^n
- (1-2*n)/(3+2*n)
- (1+2^n)/(3+2^n)
- (-1+2*n)/(3+2*n)
- (1+2*n)/(3-2*n)
Limit of the function (1+2*n)/(3+2*n)
The solution
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/1 + 2*n\ lim |-------| n->oo\3 + 2*n/
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right)$$
Limit((1 + 2*n)/(3 + 2*n), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{2 + \frac{1}{n}}{2 + \frac{3}{n}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{2 + \frac{1}{n}}{2 + \frac{3}{n}}\right) = \lim_{u \to 0^+}\left(\frac{u + 2}{3 u + 2}\right)$$
=
$$\frac{0 + 2}{3 \cdot 0 + 2} = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right) = 1$$
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{2 + \frac{1}{n}}{2 + \frac{3}{n}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{2 + \frac{1}{n}}{2 + \frac{3}{n}}\right) = \lim_{u \to 0^+}\left(\frac{u + 2}{3 u + 2}\right)$$
=
$$\frac{0 + 2}{3 \cdot 0 + 2} = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right) = 1$$
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
$$\lim_{n \to \infty}\left(2 n + 3\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(2 n + 1\right)}{\frac{d}{d n} \left(2 n + 3\right)}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(2 n + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(2 n + 3\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(2 n + 1\right)}{\frac{d}{d n} \left(2 n + 3\right)}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(\frac{2 n + 1}{2 n + 3}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{2 n + 1}{2 n + 3}\right) = \frac{1}{3}$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{2 n + 1}{2 n + 3}\right) = \frac{1}{3}$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{2 n + 1}{2 n + 3}\right) = \frac{3}{5}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{2 n + 1}{2 n + 3}\right) = \frac{3}{5}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{2 n + 1}{2 n + 3}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{2 n + 1}{2 n + 3}\right) = \frac{1}{3}$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{2 n + 1}{2 n + 3}\right) = \frac{1}{3}$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{2 n + 1}{2 n + 3}\right) = \frac{3}{5}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{2 n + 1}{2 n + 3}\right) = \frac{3}{5}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{2 n + 1}{2 n + 3}\right) = 1$$
More at n→-oo
The graph