Question

limit of (3x^4 - 4x +3)/(2x^2-5x^4+2) as x approaches infinity

Answer 1

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

Answer 2

Don't forget to put the leading term in the proper order, also, you may want to move that negative outside the fraction:

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

Answer 3

In this case, the leading term in the numerator is less than the leading term in the denominator.

Answer 4

I just wanted to post the raw problem before doing anything myself.

Answer 5

So the answer would be \[\frac{ -3 }{ 5 }\]

Answer 6

\[\lim_{x \rightarrow \infty} \frac{ 3x^4-4x+3 }{ 2x^2-5x^4+2 }=\lim_{x \rightarrow \infty}-\frac{3x^4-4x+3}{5x^4-2x^2-2}\]
And CORRECT

Answer 7

Is there no way to solve it out algebraically?

Answer 8

Good question...maybe through factoring?

Answer 9

I wonder if @Hero would know...

Answer 10

But I've been trying to factor. I can't find a way at all. I tried grouping and everything.

Answer 11

you don't solve it by factoring

Answer 12

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.

Answer 13

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?

Answer 14

I see what you're saying.