Question

I'm trying to find the limit when x->(-)infinity of (sqrt(x^6+3x^5+1)) / (7x^3+2x^5)

what do i do with the sqrt in the numerator?

is the answer infinity or sqrt(1)/7?

Answer 1

\[\sqrt(x^6+3x^5+1)/(7x^3+2x^5)\]

Answer 2

\[\lim_{x \rightarrow -\infty} \sqrt{x^6+3x^5+1}/(7x^3+2x^5)\]

Answer 3

\[\lim\limits_{x\rightarrow-∞} \frac{ \sqrt {{x^6+3x^5+1}}} {7x^3+2x^5 }\]

Answer 4

does the \[\sqrt{x^6}\] turn into \[x^3\]?

Answer 5

in the limit that x is large and negative the important terms are the \[\sqrt {x^6}\text{, and the }2x^5\]

Answer 6

but does the sqrt simplify to an x^3 or stay to the 6th power?

Answer 7

x^3 / 2x^5 ->low degree/high degree -> lim=0

Answer 8

or 6th power / 5th power -> high/low -> lim=infinity

Answer 9

yeah your right the \[√{x^6} \quad\text {would become}\quad x^3\] however x is negative so you cant do that

Answer 10

hmmm, so does the answer not exist?

Answer 11

um i am still trying to figure it out

Answer 12

okiedokie. take your time, this one has had me stumped for a long while...

Answer 13

i am just gonna use L'Hôpital's rule and see what happens

Answer 14

factor out an x^6 on the top and x^5 on the bottom

Answer 15

\[\frac{ \sqrt {{x^6+3x^5+1}}} {7x^3+2x^5 }\]
\[=\frac{ \sqrt {{1+(3/x)+(1/x^6))}}} {x^5((7/x^2)+2)}\]
\[=\frac{ |x^3|\sqrt {{x^6(1+(3/x)+(1/x^6))}}} {x^5((7/x^2)+2) }\]
=...

Answer 16

limit will be zero

Answer 17

oops typo...

Answer 18

\[=\frac{ |x^3|\sqrt {{1+(3/x)+(1/x^6)}}} {x^5((7/x^2)+2) }\]

Answer 19

do not attempt to use L'Hôpital's rule, you wil run out of paper


Zarkon has correct answer

Answer 20

oh, ok. So, when the x^6 gets pulled out it becomes an absolute value?

Answer 21

\[\sqrt{x^6}=|x^3|\]

Answer 22

thanks for going through the trouble @UnkleRhaukus!

Answer 23

and you too Zarkon!

Answer 24

ah that is what to do if you have a negative index
im glad learn something

(thanks zarkon
and thanks
hicksonm