0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Divide numerator and denominator by $x^2$. What do you get? $\endgroup$ – Daniel Fischer Mar 28 '14 at 12:57
  • $\begingroup$ 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 ? $\endgroup$ – Shuzheng Mar 28 '14 at 13:11
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It it enough to say the limit is obvious ? Or should you always prove it formally ? (I'm new to the subject) $\endgroup$ – Shuzheng Mar 28 '14 at 13:13
  • $\begingroup$ 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? $\endgroup$ – 5xum Mar 28 '14 at 13:17
2
$\begingroup$

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 $-\frac 12$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.