Question

limit (sqrt(x^4 + x^2 - 5) - x^2) as x -> inf

Wolfram Alpha says this is 1/2, but I can only figure it to be 0. Where am I wrong?

Answer 1

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

Answer 2

I divided the stuff in the square root by \[x^4\] and then simplified to be \[\lim_{x \rightarrow \infty} x^2 \times \sqrt{1 + \frac{ 1 }{ x^2 } + \frac{ 5 }{ x^4 }} - x^2\]

Answer 3

this becomes \[\lim_{x \rightarrow \infty} x^2 \times (\sqrt{1} - 1)\]

Answer 4

(note that above I went ahead and substituted in infinity for all x's but the x^2

Answer 5

Anyway, that should make the answer 0, but Wolfram Alpha says it is 1/2
So where am I wrong?

Answer 6

I think I found my own error. I'm assuming that \[\sqrt{1} - 1\] is 0, but actually the square root could be positive or negative, so you would get either -2 or 0, which is why I'm getting the problem wrong. it appears Math is saying "don't do it this way".

I would appreciate anyone else's take on the problem though, please.

Answer 7

Your error is that you're assuming

\[x^{2}(\sqrt{1+ \epsilon}-1)\]

goes to zero if epsilon goes to zero. You're not allowed to make that assumption because x^2 goes to infinity. I don't know how to solve this though.

Answer 8

\[\sqrt{x^4 + x^2 - 5} - x^2\]

\[=\sqrt{x^4 + x^2 - 5} - x^2\frac{\sqrt{x^4 + x^2 - 5} + x^2}{\sqrt{x^4 + x^2 - 5} + x^2}\]

\[=\frac{x^4 + x^2 - 5 - x^4}{\sqrt{x^4 + x^2 - 5} + x^2}=\cdots\]

Answer 9

\[=\frac{x^2 - 5 }{x^2\left(\sqrt{1 + \frac{1}{x^2} - \frac{5}{x^4}} + 1\right)}\]

\[=\frac{1 - \frac{5}{x^2} }{\sqrt{1 + \frac{1}{x^2} - \frac{5}{x^4}} + 1}\]

let \(x\to\infty\)

\[=\frac{1 - 0 }{\sqrt{1 + 0- 0} + 1}=\frac{1}{1+1}=\frac{1}{2}\]