Mathochism: It’s the little things

One woman’s attempt to revisit the math that plagued her in school. But can determination make up for 25 years of math neglect?

The test is Wednesday, and I’m trying not to let it freak me out.

So far, the lectures have been comprehensible, and the material feels strangely intuitive. I’ve been mostly in the math zone with homework, getting most problems right right away. However, I am perplexed by the fact that, as I ascend into higher, more abstract math, I get derailed by the basics.

For example — just a week or so ago, I was befuddled by the simplest absolute value problems. This week, I suddenly forgot how to add. I also often overlook negative signs, which changes the answer completely.

Thankfully, we have moved past decimals. Goodness knows what horrors I would wreak by placing that tiny dot in the wrong place!

That is not to say, of course, that I am mastering all of the new material. We had several sandwich theorem problems on the homework, and I came across one that was downright indigestible (sorry, I’m not allowed to pun much in my professional writing, so I have to get it out here).

I wasn’t the only one having trouble, which is gratifying, but the Calc Professor wouldn’t help me or anyone else solve it when we asked. He told me to think about it some more. I have! He told me to come to office hours Wednesday before the test. I will!

The issue is that in the meantime, this problem is starting to affect me in much the way a grilled panino did four years ago. Said panino was so hard I broke a molar on it. I had to get a crown, which cost me $500 even with dental insurance. I am NOT fond of that panino.

So … kind mathematicians who read this blog, I need your help. What the heck do I do with the problem below?

|x|/square root of x^4 + x^2 + 4

With some hints and a lot of sweat, I came up with this:

0 < |x|/square root of x^4 + x^2 + 4 < |x|

Fantastic! Now what? Should I do some algebra magic on the function in the middle? Say take the absolute value off x, and rationalize the denominator?

The equation should then look like this:
x (square root x^4 + x^2 + 4) / x^4 + x^2 + 4

I am stuck there. I am not even sure all of the above was necessary.

Ugh. I can feel my molar aching. And I just know calc professor is going to have something equally evil on the test. Maybe the other problems won’t be so bad?

All text copyrighted by A.K. Whitney, and cannot be used without permission.

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15 Responses to “Mathochism: It’s the little things”


  1. 1 PriestFriend March 5, 2012 at 6:57 pm

    Kind English major says, “I support your efforts wholeheartedly… to the nth degree, if you will.” Can’t. Look. at. math. equations. myself.

  2. 2 akwhitney March 5, 2012 at 7:11 pm

    It’s okay. It’s over. Now, go have a nice cup of tea. Tea fixes everything.

  3. 3 Trold March 5, 2012 at 8:13 pm

    I think you’re asking about this question as x tends to infinity? And It’s all of x4+x2+4 under the square-root sign, right? Anyway, here’s one suggestion.

    A lot of your algebra woes here come from the fact that x4+x2+4 doesn’t factor. Can you find useful information by comparing your function to similar ones with a friendlier polynomial in the denominator? For instance, would it be any easier to find the limit of |x|/sqrt(x4+4×2+4), which is (greater/less) than |x|/sqrt(x4+x2+4)? If it is, you could maybe find another similar looking function bigger than |x|/sqrt(x4+x2+4) which has the same limit as x tends to infinity and then use the squeeze theorem.

    There isn’t any rule in the squeeze theorem that the functions you use to make the sandwich have to come from the function whose limit you’re trying to evaluate. It’s totally fair to just sort of pluck them out of thin air, so long as one is reliably larger, one’s reliably smaller, and they both have the same limit at the point where you’re trying to evaluate the limit.

    Good luck!

  4. 4 Trold March 5, 2012 at 8:14 pm

    Ugh. WordPress ate my fancy formatting. Please put “^”s where needed to make exponents.

  5. 5 Olivia March 5, 2012 at 8:18 pm

    I’m with PriestFriend on that one, and I hear you on forgetting how to add. I attempted to add two three-digit numbers together the other day and screwed up carrying over. I blame it on lack of sleep!

  6. 6 L March 5, 2012 at 9:00 pm

    You have |x|/square root of x^4 + x^2 + 4 < |x| so you're halfway there! You just need to find a tighter lower bound than zero. Think about how you found the upper bound and try tweaking that approach so that it gives you a lower bound. Then you can apply the squeeze theorem.

  7. 7 akwhitney March 5, 2012 at 9:14 pm

    Thanks L and Troid! Sadly, in spite of your kind explanations, I am even more confuzzled than before.

  8. 8 Dave W. March 6, 2012 at 1:00 am

    Well, you might want to consider that |x| = sqrt(x^2) (for all real x). Also, the quotient of two square roots is the square root of the quotient: [sqrt(p)/sqrt(q) = sqrt(p/q)]. And x^4 < x^4 + x^2 + 4, so sqrt(x^4) 1/sqrt(x^4 + x^2 + 4). Is that enough to work with?

    Another general point: when working with limits involving absolute values, the only absolute values you really need to worry about are those where the content of the absolute value switches signs close to the limit. For any positive value of x, |x| = x, so if you are evaluating a positive limit, you can just substitute x for |x|, since you only need to worry about what happens for positive x. Likewise for negative limits, you could substitute -x for |x|, since you are only worried about what happens to the function for negative x. Only if you are evaluating the limit as x->0 do you really need to keep the |x| around to be careful.

  9. 9 Dave W. March 6, 2012 at 1:04 am

    Killed by html. Let’s try that again:
    x^4 &lt x^4 + x^2 + 4, so sqrt(x^4) &lt sqrt(x^4 + x^2 + 4) and therefore 1/sqrt(x^4) &gt 1/sqrt(x^4 + x^2 + 4).

  10. 10 L March 6, 2012 at 8:11 am

    Lets call the function that we’re dealing with g(x). To apply the squeeze theorem you need to find f(x) and h(x) such that
    1. f(x) < g(x)<h(x) *
    2. the limit of f(x) equals the limit of h(x).

    You've found out that 0 < g(x) < |x|. If you choose f(x)=0 and h(x)=|x|, you've satisfied the first requirement. However this choice of f(x) and h(x) fails the second requirement. Your choice of h(x) is fine but you need to find a new f(x).

    Trold explained how to go about finding f(x). Take a look at the function that s/he mentioned- it's a great choice for f(x). Hope this helps!

    *< should be 'less than or equal' but there's not a clean way to type that.

  11. 11 Antonia March 6, 2012 at 9:38 am

    Just to double-check, could you write out the whole problem? Like, is it:

    lim(x->a)(|x|/square root of x^4 + x^2 + 4)? If so, what is the value of a given in the problem?

  12. 12 akwhitney March 6, 2012 at 9:48 am

    Okay, here’s the actual problem. The polynomial is different — I was hoping to get some ideas for solving this beast, so I tweaked it a bit. Apparently, that was a tactical error. Eep.

    Use the sandwich theorem to verify the limit:
    lim as x approaches 0 |x|/square root x^4 + 4x^2 + 7 = 0
    (Hint: Use f(x) = 0 and g(x) = |x|)

  13. 13 Dave W. March 6, 2012 at 11:05 am

    Yeah, changing the problem can make a difference, but the bigger issue is that you never specified what x was approaching in the original problem. Now that we know that we want the limit at 0 (rather than infinity or something else), the problem gets easier.

    Both f and g have limits of 0 as x approaches 0, so all you need to do is show the inequalities and then you can apply the sandwich theorem. Assuming that the denominator of your original function is sqrt(x^4 + 4x^2 + 7) (i.e., that all that polynomial is inside the square root sign), then you know that the denominator is always greater than or equal to sqrt(7), because the two terms involving x are always non-negative. So your function is always at most |x| / sqrt(7), which is less than or equal to g(x). By the same token, your numerator is always greater than or equal to 0, and you are dividing by something strictly positive (which you know to be at least sqrt(7)), so the fraction is always going to be greater than or equal to 0, which is f(x). Now go ahead and sandwich away.

  14. 14 L March 6, 2012 at 11:59 am

    I was assuming it was as x approaches positive infinity. Ignore everything I said above.

  15. 15 akwhitney March 6, 2012 at 12:01 pm

    Thanks so much for being willing to help! My bad for not specifying. I’m still getting the hang of this.


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