Question

Evaluate the limit, please!
http://prntscr.com/d5e2x4

Answer 1

hint: divide each term by \(\Large x^3\)

Answer 2

Yes, then?

Answer 3

What do you notice about the numerator? What happens when x approaches infinity?

Answer 4

hint:

After dividing each term by x^3, the numerator goes from
\[\Large \lim_{x \to \infty} (-8x^2+5x)\]
to
\[\Large \lim_{x \to \infty} \left(-\frac{8}{x}+\frac{5}{x^2}\right)\]

Answer 5

it will be like that:
\[\Large \frac{ -8x^2-5x }{ -8x^3+7x^2-5x-4 }=\frac{ x^3[-\frac{ 8x^2}{ x^3 }-\frac{ 5x }{ x^3 }] }{ x^3[-\frac{ 8x^3 }{ x^3 }+\frac{ 7x^2 }{ x^3 }-\frac{ 5x }{ x^3 }-\frac{ 4 }{ x^3 }]}\]

Answer 6

+5x up top, not -5x

but you have the right idea

Answer 7

focus on the individual fractions

each fraction with an x in the denominator will approach 0 as x gets larger and larger

so you'll end up with 0 up top and -8 down below

Answer 8

again the -5x should be a +5x

Answer 9

\[\large \lim_{x \rightarrow \infty}\frac{ x^3[-\frac{ 8x^2}{ x^3 }+\frac{ 5x }{ x^3 }] }{ x^3[-\frac{ 8x^3 }{ x^3 }+\frac{ 7x^2 }{ x^3 }-\frac{ 5x }{ x^3 }-\frac{ 4 }{ x^3 }]}=\lim_{x \rightarrow \infty}\frac{-\frac{ 8 }{ x }+\frac{ 5 }{ x^2 } }{ -8+\frac{ 7 }{ x }-\frac{ 5 }{ x^2 }-\frac{ 4 }{ x^3 } }=\]

Answer 10

the proper way is to divide each term by the highest power, then use the limit
\[\lim_{x \rightarrow \pm \infty} \frac{ 1 }{ x^n } = 0 ~whenever ~n \ge 1\]

However after trying this several times you realise a shortcut
If the rational function is
\[\frac{ P(x) }{ Q(x) }\]
and degree(P) < deg(Q), then the limit will be 0 always.

If degree(P)=degree(Q) then the limit is the quotient of the leading coefficients.
if degree(P) > degree(Q) then your limit does not exist and you will have an oblique asymptote on your graph.

Answer 11

\[\large=\lim_{x \rightarrow \infty}\frac{-\frac{ 8 }{ x }-\frac{ 5 }{ x^2 } }{ -8+\frac{ 7 }{ x }-\frac{ 5 }{ x^2 }-\frac{ 4 }{ x^3 } }=\Large \frac{ 0-0 }{ -8+0-0-0 }=\color{MediumSpringGreen }{\frac{ 0 }{ -8 }}=\Huge\color{Lime }{0}\]

I think it is the right answer here!

Answer 12

@jim_thompson5910
I think I got it right. Don't I?

Answer 13

Anyway, Thank you very much.

Answer 14

yes the final answer is 0

the denominator is growing faster than the numerator, so this is an informal alternative to get the same answer

Answer 15

"an informal alternative to get the same answer"??

Answer 16

yes the idea that if the denominator has a larger degree, then the limit at infinity is 0

Answer 17

Yes I think I got it