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(-5+7*n)/(2+n)
Limit of the function/ (-5+7*n)/(2+n)

Limit of the function (-5+7*n)/(2+n)

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     /-5 + 7*n\
 lim |--------|
n->oo\ 2 + n  /
lim⁡n→∞(7n−5n+2)\lim_{n \to \infty}\left(\frac{7 n - 5}{n + 2}\right) 
Limit((-5 + 7*n)/(2 + n), n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞(7n−5n+2)\lim_{n \to \infty}\left(\frac{7 n - 5}{n + 2}\right)
Let's divide numerator and denominator by n:
lim⁡n→∞(7n−5n+2)\lim_{n \to \infty}\left(\frac{7 n - 5}{n + 2}\right) =
lim⁡n→∞(7−5n1+2n)\lim_{n \to \infty}\left(\frac{7 - \frac{5}{n}}{1 + \frac{2}{n}}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(7−5n1+2n)=lim⁡u→0+(7−5u2u+1)\lim_{n \to \infty}\left(\frac{7 - \frac{5}{n}}{1 + \frac{2}{n}}\right) = \lim_{u \to 0^+}\left(\frac{7 - 5 u}{2 u + 1}\right)
=
7−00⋅2+1=7\frac{7 - 0}{0 \cdot 2 + 1} = 7

The final answer:
lim⁡n→∞(7n−5n+2)=7\lim_{n \to \infty}\left(\frac{7 n - 5}{n + 2}\right) = 7
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
lim⁡n→∞(7n−5)=∞\lim_{n \to \infty}\left(7 n - 5\right) = \infty
and limit for the denominator is
lim⁡n→∞(n+2)=∞\lim_{n \to \infty}\left(n + 2\right) = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡n→∞(7n−5n+2)\lim_{n \to \infty}\left(\frac{7 n - 5}{n + 2}\right)
=
lim⁡n→∞(ddn(7n−5)ddn(n+2))\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(7 n - 5\right)}{\frac{d}{d n} \left(n + 2\right)}\right)
=
lim⁡n→∞7\lim_{n \to \infty} 7
=
lim⁡n→∞7\lim_{n \to \infty} 7
=
77
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
02468-8-6-4-2-1010-500500
Plot the graph
Other limits n→0, -oo, +oo, 1
lim⁡n→∞(7n−5n+2)=7\lim_{n \to \infty}\left(\frac{7 n - 5}{n + 2}\right) = 7
lim⁡n→0−(7n−5n+2)=−52\lim_{n \to 0^-}\left(\frac{7 n - 5}{n + 2}\right) = - \frac{5}{2}
More at n→0 from the left
lim⁡n→0+(7n−5n+2)=−52\lim_{n \to 0^+}\left(\frac{7 n - 5}{n + 2}\right) = - \frac{5}{2}
More at n→0 from the right
lim⁡n→1−(7n−5n+2)=23\lim_{n \to 1^-}\left(\frac{7 n - 5}{n + 2}\right) = \frac{2}{3}
More at n→1 from the left
lim⁡n→1+(7n−5n+2)=23\lim_{n \to 1^+}\left(\frac{7 n - 5}{n + 2}\right) = \frac{2}{3}
More at n→1 from the right
lim⁡n→−∞(7n−5n+2)=7\lim_{n \to -\infty}\left(\frac{7 n - 5}{n + 2}\right) = 7
More at n→-oo
Rapid solution [src] 
7
77
Expand and simplify
The graph
Limit of the function (-5+7*n)/(2+n)
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