anonymous
  • anonymous
What is the limit as x approaches infinity for f(x)=4x^2+8x/x^3+7x^2-x-9?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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math&ing001
  • math&ing001
When calculating limits near infinity we only take in consideration the highest degree term: \[\lim_{x \rightarrow \infty} \frac{ 4x ^{2}+8x }{ x ^{3}+7x ^{2}-x-9 }=\lim_{x \rightarrow \infty} \frac{ 4x ^{2}}{ x ^{3}}\]
math&ing001
  • math&ing001
@Camila1315 can you do it now ?
anonymous
  • anonymous
Yes the limit would be infinity right?

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math&ing001
  • math&ing001
Nope, how did you infinity ?
math&ing001
  • math&ing001
did you get*
anonymous
  • anonymous
When I graphed it I saw the when x kept approaching infinity, the y value was coming from infinity.
math&ing001
  • math&ing001
No need to graph it, just go from here \(\lim_{x \rightarrow \infty} \frac{ 4x ^{2} }{ x ^{3} }\) and simplify.
anonymous
  • anonymous
Oh is it 4?
math&ing001
  • math&ing001
\[\lim_{x \rightarrow \infty} \frac{ 4x ^{2} }{ x ^{3} }=\lim_{x \rightarrow \infty} \frac{ 4 }{ x }\] Do you know the limit of 1/x near infinity ?
anonymous
  • anonymous
Im so bad with limits I thought the limit there would be infinity
math&ing001
  • math&ing001
O.K here's a graph that shows you how the 1/x function acts near infinity : |dw:1433595410014:dw|
anonymous
  • anonymous
Oh so you have to look at x when it is approaching infinity?
math&ing001
  • math&ing001
Yeah, that's how you get the limit from the graph. Can you figure out the limit now ?
anonymous
  • anonymous
Yeah so the limit would be 0
math&ing001
  • math&ing001
Yep !

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