Question

@Michele_Laino

Answer 1

from your data, I can rewrite your function as follows:
\[f(x)=\frac{ x ^{3}(2+9/x ^{3} )}{ x ^{4}(1-81/x ^{4}) }=\]
\[=\frac{ 1 }{ x }\frac{ 2+\frac{ 9 }{ x ^{3} } }{ 1-\frac{ 81 }{ x ^{4} } }=\]

Answer 2

now, as before, when x--->+infinity, 1/x^3--->?

Answer 3

please try!

Answer 4

noo, please if x--->+infinity, then 1/x^3--->0

Answer 5

ohhh

Answer 6

also 1/x^4--->0,so we can write:
\[f(x)\rightarrow 0*\frac{ 2+0 }{ 1-0 }=0\]
because, when x--->infinity, 1/x--->0

Answer 7

So the first option is the right answer!

Answer 8

i have one more(:

Answer 9

of course we have:
f(x)--->0also when x-->-infinity
so we can write:
\[\lim _{x \rightarrow \pm \infty} f(x)=0\]

Answer 10

derivative of:
\[f(x)=3xe ^{-x}\]
is:?

Answer 11

-3e^-x(x-1)

Answer 12

perfect!, when is it zero?

Answer 13

I remmber you, please that function e^-x, never reaches zero, more precisely, it reaches zero at x--->+infinity: