Question

@Michele_Laino

Answer 1

hint: evaluate this limit, please:
\[\lim _{x \rightarrow \infty}\frac{ e ^{x+3} }{ e ^{x} }=...\]

Answer 2

another hint:
please note that:
\[\frac{ e ^{x+3} }{ e ^{x} }=e ^{3}\]
so?

Answer 3

it is B!

Answer 4

no, because:
\[\lim _{x \rightarrow +\infty}\frac{ e ^{3x} }{ e ^{x} }=\lim _{x \rightarrow +\infty}e ^{2x}=+\infty\]

Answer 5

oh so it is C?

Answer 6

no, because:
\[\lim _{x \rightarrow +\infty}\frac{ e ^{2x} }{ e ^{x} }=\lim _{x \rightarrow + \infty}e ^{x}=+\infty\]

Answer 7

yes!, because:
\[\lim _{x \rightarrow +\infty}\frac{ e ^{x+3} }{ e ^{x} }=e ^{3}\neq 0\]

Answer 8

hint:
\[\frac{ x+sinx }{ x }=1+\frac{ \sin x }{ x }\]
so?

Answer 9

they are the same rate!

Answer 10

yes, because if x>0, since x---> +infinity, we have:
\[-\frac{ 1 }{ x }\le \frac{ \sin x }{ x }\le \frac{ 1 }{ x }\]
since:
\[-\frac{ 1 }{ x }\rightarrow 0, \frac{ 1 }{ x }\rightarrow 0\]
when x--->+infty,
we have:
\[\lim _{x \rightarrow +\infty}\frac{ \sin x }{ x }=0\]
and:
\[\lim _{x \rightarrow +\infty}\left( \frac{ x+ \sin x }{ x } \right)=0\]

Answer 11

of course, the same result we will get in the case x--->-infinity

Answer 12

namely:
\[x \rightarrow -\infty\]

Answer 13

so it is C (:

Answer 14

oops...
\[\lim _{x \rightarrow +\infty}\left( \frac{ x+ \sin x }{ x } \right)=1\]

Answer 15

wait is it c?

Answer 16

yes! it is C

Answer 17

thank you !!!!

Answer 18

thank you! :)

Answer 19

thank you !!!!