# Is it possible to apply L'Hôpital's rule to $\lim_{x \to \infty} \left(x^2 - \sqrt{x^4 + 4}\right)$?

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.

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

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 $$2\lim_{y\to 0^+}{1-\sqrt{1+y^2}\over y}\underset{H}{=}2 \lim_{y\to 0^+}{-y\over\sqrt{1+y^2}}=0$$

• Thank you, this is what I was looking for. Very insightful. Commented Apr 11 at 14:08
• 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. Commented Apr 11 at 14:13
• 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: $\lim_{x \to \infty} \frac{1 - \sqrt{1 + 4x^{-4}}}{x^{-2}} \stackrel{\text{H}}{=} \lim_{x \to \infty} \frac{-\frac{1}{2} \cdot \sqrt{1 + 4x^{-4}} \cdot (-16x^{-5})}{-2x^{-3}}$ Commented Apr 11 at 14:15
• You are right. I will re-edit. Fortunately it didn't affect the limit. Commented Apr 11 at 14:18

Nope. Its not a fraction.

But, the standard conjugating works.

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