anonymous
  • anonymous
limit of (3x^4 - 4x +3)/(2x^2-5x^4+2) as x approaches infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\lim_{x \rightarrow \infty} \frac{ 3x^4-4x+3 }{ 2x^2-5x^4+2 }\]
Hero
  • Hero
Don't forget to put the leading term in the proper order, also, you may want to move that negative outside the fraction: \[\lim_{x \rightarrow \infty} -\frac{ 3x^4-4x+3 }{5x^4-2x^2-2 }\]
Hero
  • Hero
In this case, the leading term in the numerator is less than the leading term in the denominator.

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anonymous
  • anonymous
I just wanted to post the raw problem before doing anything myself.
anonymous
  • anonymous
So the answer would be \[\frac{ -3 }{ 5 }\]
anonymous
  • anonymous
\[\lim_{x \rightarrow \infty} \frac{ 3x^4-4x+3 }{ 2x^2-5x^4+2 }=\lim_{x \rightarrow \infty}-\frac{3x^4-4x+3}{5x^4-2x^2-2}\] And CORRECT
anonymous
  • anonymous
Is there no way to solve it out algebraically?
anonymous
  • anonymous
Good question...maybe through factoring?
anonymous
  • anonymous
I wonder if @Hero would know...
anonymous
  • anonymous
But I've been trying to factor. I can't find a way at all. I tried grouping and everything.
Hero
  • Hero
you don't solve it by factoring
anonymous
  • anonymous
I'm just going to use the way that I know and tell my teacher that I couldn't figure out how to solve it algebraically. Thanks for your help @KeithAfasCalcLover and @Hero.
anonymous
  • anonymous
But you see, the problem with that is that \[\lim_{x\rightarrow\infty}{-\frac{3x^4-4x+3}{5x^4-2x^2-2}}=\lim_{x\rightarrow\infty}{-\left(\frac{3x^4}{5x^4-2x^2-2}+\frac{-4x}{5x^4-2x^2-2}+\frac{3}{5x^4-2x^2-2}\right)}\] \[=\lim_{x\rightarrow\infty}{\frac{-3x^4}{5x^4-2x^2-2}}+\lim_{x\rightarrow\infty}{\frac{4x}{5x^4-2x^2-2}}+\lim_{x\rightarrow\infty}{\frac{-3}{5x^4-2x^2-2}}\] So how each of these cope with that rule?
anonymous
  • anonymous
I see what you're saying.

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