mathcalculus
HELP: evaluating limits to infinity by dividing x
(ATTACHED BELOW)



This Question is Closed

mathcalculus
Best Response
You've already chosen the best response.
1

mathcalculus
Best Response
You've already chosen the best response.
1
i got 3/5 but it's wrong.

mathcalculus
Best Response
You've already chosen the best response.
1
@kropot72 do you know this?

mathcalculus
Best Response
You've already chosen the best response.
1
@satellite73 any ideas?

hihihii
Best Response
You've already chosen the best response.
0
0

hihihii
Best Response
You've already chosen the best response.
0
if the denominator grows bigger than the numerator, then the limit is 0

mathcalculus
Best Response
You've already chosen the best response.
1
oh so we don;t need to divide by x? or do anything to it?

hihihii
Best Response
You've already chosen the best response.
0
nope!

mathcalculus
Best Response
You've already chosen the best response.
1
thanks! :) @hihihii

hihihii
Best Response
You've already chosen the best response.
0
no problem!

mathstudent55
Best Response
You've already chosen the best response.
1
\[\lim_{x \rightarrow \infty} \dfrac{3x + 2}{3x^2  5x + 8}\]
\[ =\lim_{x \rightarrow \infty} \dfrac{ x^2 \left( \frac{3}{x} + \frac{2}{x^2} \right) } {x^2 \left( 3  \frac{5}{x} + \frac{8}{x^2} \right) }\]
\[ = \lim_{x \rightarrow \infty} \dfrac{ \frac{3}{x} + \frac{2}{x^2} } { 3  \frac{5}{x} + \frac{8}{x^2} } \]
\[ = \dfrac{ 0 + 0} { 3  0 + 0 } \]
\(= 0\)

mathcalculus
Best Response
You've already chosen the best response.
1
thanks u @mathstudent55

mathstudent55
Best Response
You've already chosen the best response.
1
You're welcome.