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I evaluated a limit similar to this one for homework by rationalizing the numerator then dividing by the largest-degreed x term in the denominator:

$$\lim_{x \to \infty} \left(x^2 - \sqrt{x^4 + 4}\right)$$

However, I noticed that evaluating naively would lead to $\infty - \infty$, an indeterminate form, so I was wondering if it was possible to use L'Hôpital's rule for it?

I tried a few times but couldn't get the right answer. Maybe it's not possible? I kept ending up with $\infty$, which was incorrect.

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    $\begingroup$ I would evaluate it as $x^2\left(1-\sqrt{1+\dfrac4{x^4}}\right)$ and use binomial series for $\left(1+\dfrac4{x^4}\right)^{1/2}$ $\endgroup$
    – J. W. Tanner
    Commented Apr 11 at 3:00
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    $\begingroup$ @J.W.Tanner FYI, I believe it's somewhat simpler & easier to instead rationalize the limit expression by multiplying it by $\frac{x^2+\sqrt{x^4+4}}{x^2+\sqrt{x^4+4}}$. $\endgroup$
    – John Omielan
    Commented Apr 11 at 3:02
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    $\begingroup$ Just $\left(1+\dfrac4{x^4}\right)^{1/2}=1+\dfrac2{x^4}\cdots$ $\endgroup$
    – J. W. Tanner
    Commented Apr 11 at 3:10

2 Answers 2

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It is possible to apply the rule, by brutal force (not recommended). $$\lim_{x\to \infty}[x^2-\sqrt{x^4+4}]=\lim_{x\to \infty} {1-\sqrt{1+4x^{-4}}\over x^{-2}}\\ \underset{H}{=}-\lim_{x\to\infty}{{{16x^{-5}\over2\sqrt{1+4x^{-4}}}}\over 2x^{-3}}=-\lim_{x\to\infty}{4x^{-2}\over \sqrt{1+4x^{-4}}}=0$$

Remark The calculations simplify substantially if we subsitute $y=2x^{-2}$ in the second expression. In this way we get $$2\lim_{y\to 0^+}{1-\sqrt{1+y^2}\over y}\underset{H}{=}2 \lim_{y\to 0^+}{-y\over\sqrt{1+y^2}}=0$$

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  • $\begingroup$ Thank you, this is what I was looking for. Very insightful. $\endgroup$
    – rainbridge
    Commented Apr 11 at 14:08
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    $\begingroup$ Thanks for accepting. Personally before applying l'Hopital I would subsitute $y=x^{-2},$ so $y\to 0^+$ and then apply l'Hopital. The calculations become simpler. $\endgroup$
    – Ryszard Szwarc
    Commented Apr 11 at 14:13
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    $\begingroup$ Also, following along, I'm getting the negative of what you have after applying L'Hôpital's Rule — am I missing something? I saw three negatives: $\lim_{x \to \infty} \frac{1 - \sqrt{1 + 4x^{-4}}}{x^{-2}} \stackrel{\text{H}}{=} \lim_{x \to \infty} \frac{-\frac{1}{2} \cdot \sqrt{1 + 4x^{-4}} \cdot (-16x^{-5})}{-2x^{-3}} $ $\endgroup$
    – rainbridge
    Commented Apr 11 at 14:15
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    $\begingroup$ You are right. I will re-edit. Fortunately it didn't affect the limit. $\endgroup$
    – Ryszard Szwarc
    Commented Apr 11 at 14:18
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Nope. Its not a fraction.

But, the standard conjugating works.

$\begin{array}\\ \left(x^2 - \sqrt{x^4 + 4}\right) &=\left(x^2 - \sqrt{x^4 + 4}\right)\dfrac{x^2 + \sqrt{x^4 + 4}}{x^2 + \sqrt{x^4 + 4}}\\ &=\dfrac{x^4 - (x^4 + 4)}{x^2 + \sqrt{x^4 + 4}}\\ &=\dfrac{-4}{x^2 + \sqrt{x^4 + 4}}\\ &\to 0\\ \end{array} $

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