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