# Show $f(x)=\sqrt{x^4+1} - \sqrt{x^4+x^2} \rightarrow -1/2$ for $x \rightarrow \infty$, $x \in \mathbb R$.

Show $f(x)=\sqrt{x^4+1} - \sqrt{x^4+x^2} \rightarrow -1/2$ for $x \rightarrow \infty$, $x \in \mathbb R$.

I've tried $$\frac {(\sqrt{x^4+1} - \sqrt{x^4+x^2})(\sqrt{x^4+1} + \sqrt{x^4+x^2})}{\sqrt{x^4+1} + \sqrt{x^4+x^2} } = \frac {1-x^2} {\sqrt{x^4+1} + \sqrt{x^4+x^2}}$$

and $$\frac {1} {\sqrt{x^4+1} + \sqrt{x^4+x^2}} \rightarrow 0$$ so it's enough to verify $$\frac {-x^2} {\sqrt{x^4+1} + \sqrt{x^4+x^2}} \rightarrow -1/2$$.

However I'm having trouble showing this.

• Divide numerator and denominator by $x^2$. What do you get? – Daniel Fischer Mar 28 '14 at 12:57
• Thank you Daniel, I get a result from which I can verify that the limit is indeed $-1/2$. I'm new to this subject. Could you say how rigorous I should be in proving from the result that the limit is $-1/2$ ? - Should I verify this by using $\epsilon$ and $\delta$, or is it enough to verify it by inspection ? – Shuzheng Mar 28 '14 at 13:11

Yes, is enough to verify the last expression. This is because of $f$ and $g$ both have limits as $x$ approaches $\infty$, then $\lim_{x\to\infty}f(x)+g(x) = \lim_{x\to\infty}f(x) + \lim_{x\to\infty} g(x)$.
When calculating the last limit, you can simply divide both sides of the fraction by $x^2$ to get
$$\frac{-x^2}{\sqrt{x^4+1} + \sqrt{x^4+x^2}} = \frac{-1}{\sqrt{1+\frac{1}{x^4}} + \sqrt{1 + \frac{1}{x^2}}}$$ which has an obvious limit.
• Yes, of course you must prove it, but it is not hard, is it? I mean, you know what the limit of $\frac1x$ is when $x\to\infty$, don't you? – 5xum Mar 28 '14 at 13:17
In this kind of problems which want you to calculate the limit when $x \to \infty$, constants are not very important and you have to pay attention to powers of $x$ (or anything related to $x$). That makes sense because, e.g., when $x=10$, you have $x^4 = 10000$, which is way greater than $1$.
Try to convert $x^4+x^2$ to a perfect square, that is: $$x^4+x^2 +\frac 14 - \frac 14 = (x^2+\frac 12)^2 - \frac 14.$$ Now, take perfect squares out of radical. Your limit becomes $$x^2\sqrt{1+\frac{1}{x^4}} - (x^2+\frac 12)\sqrt{1+\frac{1}{(x^2+\frac 12)^2}}, \quad x \to \infty$$ You see that the phrases under radical go to $1$ as $x$ goes to $\infty$, and the answer is simply $-\frac 12$.