Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-2*sin(2*x)+sin(4*x))/(x*log(cos(6*x)))
- Limit of i*n*(1+x)/x
-
Limit of (5+(1+n)^2+3*n)/(2+n^2+3*n)
-
Limit of (4-x)/(4+x^2-5*x)
-
Identical expressions
- four *n/(- one +n)
- 4 multiply by n divide by ( minus 1 plus n)
- four multiply by n divide by ( minus one plus n)
- 4n/(-1+n)
- 4n/-1+n
- 4*n divide by (-1+n)
-
Similar expressions
- (5+4*n)/(-1+n+n^2)
- 4*n/(-1-n)
- 4*n/(1+n)
Limit of the function 4*n/(-1+n)
The solution
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/ 4*n \ lim |------| n->oo\-1 + n/
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$
Limit((4*n)/(-1 + n), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{4}{1 - \frac{1}{n}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{4}{1 - \frac{1}{n}}\right) = \lim_{u \to 0^+}\left(\frac{4}{1 - u}\right)$$
=
$$\frac{4}{1 - 0} = 4$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right) = 4$$
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$
Let's divide numerator and denominator by n:
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{4}{1 - \frac{1}{n}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{4}{1 - \frac{1}{n}}\right) = \lim_{u \to 0^+}\left(\frac{4}{1 - u}\right)$$
=
$$\frac{4}{1 - 0} = 4$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right) = 4$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(4 n\right) = \infty$$
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.
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 4 n}{\frac{d}{d n} \left(n - 1\right)}\right)$$
=
$$\lim_{n \to \infty} 4$$
=
$$\lim_{n \to \infty} 4$$
=
$$4$$
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
$$\lim_{n \to \infty}\left(4 n\right) = \infty$$
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.
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 4 n}{\frac{d}{d n} \left(n - 1\right)}\right)$$
=
$$\lim_{n \to \infty} 4$$
=
$$\lim_{n \to \infty} 4$$
=
$$4$$
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
$$\lim_{n \to \infty}\left(\frac{4 n}{n - 1}\right) = 4$$
$$\lim_{n \to 0^-}\left(\frac{4 n}{n - 1}\right) = 0$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{4 n}{n - 1}\right) = 0$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{4 n}{n - 1}\right) = -\infty$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{4 n}{n - 1}\right) = \infty$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{4 n}{n - 1}\right) = 4$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{4 n}{n - 1}\right) = 0$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{4 n}{n - 1}\right) = 0$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{4 n}{n - 1}\right) = -\infty$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{4 n}{n - 1}\right) = \infty$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{4 n}{n - 1}\right) = 4$$
More at n→-oo
The graph