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