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$$\lim_{x\to \infty} \frac{x^2}{\sqrt{x^4+6}} $$

Shouldn't this limit be equal to 0 ? We divide the numerator and the denominator by $x^4$ and we get $1 / x^2$ in the numerator. So when x approaches infinity, the numerator becomes 0.

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    $\begingroup$ But then the denominator will also become $0$. $\endgroup$ – Colescu Oct 5 '16 at 5:23
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    $\begingroup$ We divide the numerator and the denominator by x^4 and ... Apply the same line of thought to $x^2 / \sqrt{x^4}$. Find where it fails. Reconsider. $\endgroup$ – dxiv Oct 5 '16 at 5:28
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If you divide the numerator and denominator by $x^4$, you get $$ \lim_{x\to \infty} \frac{1/x^2}{\sqrt{(1/x^4)+(6/x^8)}} $$ Plugging in $x = \infty$, the numerator and denominator are both $0$. So it is inconclusive.

We can instead divide the numerator and denominator by $x^2$. This gives us $$ \lim_{x\to \infty} \frac{1}{\sqrt{1+(6/x^4)}} $$ and plugging in $x = \infty$ we get $\frac{1}{\sqrt{1}} = 1$, which is the right answer.

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  • $\begingroup$ You misspoke, you cannot plugin infinity, you meant the limit. $\endgroup$ – Zelos Malum Oct 5 '16 at 5:27
  • $\begingroup$ @ZelosMalum No, I did not misspeak. There is no problem plugging in $\infty$ to these expressions, or with arithmetic in general involving $\infty$ as long as you stop whenever you reach something like $\frac{\infty}{\infty}$ or $\infty - \infty$. $\endgroup$ – 6005 Oct 5 '16 at 5:29
  • $\begingroup$ Infinity is not a real number so it is as meaningful saying "plug in apple", or more mathematicly "using integer multiplication what is 7 times the imaginary unit?" It's meaningless as that unit does not belong in integers so that operation is undefined. $\endgroup$ – Zelos Malum Oct 5 '16 at 5:31
  • $\begingroup$ @ZelosMalum OK, so I'm doing something differently than what you did in your freshman calculus class -- but do you really think that's the only way of doing things? Educate yourself on the extended real numbers arithmetic. $\endgroup$ – 6005 Oct 5 '16 at 5:33
  • $\begingroup$ no need, THERE it is legitimate, not in the reals and the poster did not say he worked with the extended reals. As for wrong, only in that instance of being rigorous but the idea itself, no. $\endgroup$ – Zelos Malum Oct 5 '16 at 5:35
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multiply the numerator and denominator by $1/x^2$.

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$$\lim_{x\to \infty} = \frac{x^2}{\sqrt{x^4+6}}=\lim_{x\to \infty}\frac{1}{\sqrt{1+6/x^4}}=1 $$

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