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(3+2*x)/(1+5*x)
Limit of the function/ (3+2*x)/(1+5*x)

Limit of the function (3+2*x)/(1+5*x)

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     /3 + 2*x\
 lim |-------|
x->oo\1 + 5*x/
lim⁡x→∞(2x+35x+1)\lim_{x \to \infty}\left(\frac{2 x + 3}{5 x + 1}\right) 
Limit((3 + 2*x)/(1 + 5*x), x, oo, dir='-')
Detail solution
Let's take the limit
lim⁡x→∞(2x+35x+1)\lim_{x \to \infty}\left(\frac{2 x + 3}{5 x + 1}\right)
Let's divide numerator and denominator by x:
lim⁡x→∞(2x+35x+1)\lim_{x \to \infty}\left(\frac{2 x + 3}{5 x + 1}\right) =
lim⁡x→∞(2+3x5+1x)\lim_{x \to \infty}\left(\frac{2 + \frac{3}{x}}{5 + \frac{1}{x}}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
lim⁡x→∞(2+3x5+1x)=lim⁡u→0+(3u+2u+5)\lim_{x \to \infty}\left(\frac{2 + \frac{3}{x}}{5 + \frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 2}{u + 5}\right)
=
0⋅3+25=25\frac{0 \cdot 3 + 2}{5} = \frac{2}{5}

The final answer:
lim⁡x→∞(2x+35x+1)=25\lim_{x \to \infty}\left(\frac{2 x + 3}{5 x + 1}\right) = \frac{2}{5}
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
lim⁡x→∞(2x+3)=∞\lim_{x \to \infty}\left(2 x + 3\right) = \infty
and limit for the denominator is
lim⁡x→∞(5x+1)=∞\lim_{x \to \infty}\left(5 x + 1\right) = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→∞(2x+35x+1)\lim_{x \to \infty}\left(\frac{2 x + 3}{5 x + 1}\right)
=
lim⁡x→∞(ddx(2x+3)ddx(5x+1))\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(2 x + 3\right)}{\frac{d}{d x} \left(5 x + 1\right)}\right)
=
lim⁡x→∞25\lim_{x \to \infty} \frac{2}{5}
=
lim⁡x→∞25\lim_{x \to \infty} \frac{2}{5}
=
25\frac{2}{5}
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
02468-8-6-4-2-1010-1010
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Rapid solution [src] 
2/5
25\frac{2}{5}
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(2x+35x+1)=25\lim_{x \to \infty}\left(\frac{2 x + 3}{5 x + 1}\right) = \frac{2}{5}
lim⁡x→0−(2x+35x+1)=3\lim_{x \to 0^-}\left(\frac{2 x + 3}{5 x + 1}\right) = 3
More at x→0 from the left
lim⁡x→0+(2x+35x+1)=3\lim_{x \to 0^+}\left(\frac{2 x + 3}{5 x + 1}\right) = 3
More at x→0 from the right
lim⁡x→1−(2x+35x+1)=56\lim_{x \to 1^-}\left(\frac{2 x + 3}{5 x + 1}\right) = \frac{5}{6}
More at x→1 from the left
lim⁡x→1+(2x+35x+1)=56\lim_{x \to 1^+}\left(\frac{2 x + 3}{5 x + 1}\right) = \frac{5}{6}
More at x→1 from the right
lim⁡x→−∞(2x+35x+1)=25\lim_{x \to -\infty}\left(\frac{2 x + 3}{5 x + 1}\right) = \frac{2}{5}
More at x→-oo
The graph
Limit of the function (3+2*x)/(1+5*x)
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