0

$$\lim_{x \to 16}\; \dfrac{(4-\sqrt x)}{(16x-x^2)}$$

I tried to multiply it all by the (forgot the term) $4+\sqrt x$

but that doesn't seem to help, I get 16-16/a very negative number.

Asaf Karagila
  • 393,674

4 Answers4

3

I see you've asked a series of these questions now, so I think that a simple hint will suffice.

HINT:

$(16x - x^2) = x(16 - x) = x(4 - \sqrt x)(4 + \sqrt x)$

davidlowryduda
  • 91,687
  • I don't quit follow that order, seems like jumps after the second change. –  Aug 28 '11 at 03:00
  • @Jordan: I don't understand what you mean. Are you saying that you don't see that the second term is equal to the third term? I recommend multiplying out the 3rd term then, to see if you get the second back. I assume you know the factorization $a^2 - b^2 = (a+b)(a-b)$? – davidlowryduda Aug 28 '11 at 03:02
  • No, I forgot all that. Is this something I will need to memorize again? –  Aug 28 '11 at 03:03
  • @Jordan: Well, I used it here. I'm always reluctant to tell people to memorize things. I think I use this factorization frequently, but I use many factorizations frequently. I need to rederive many all the time to remember what they are because I hate memorizing things. So this isn't really an answer, but I do think it's useful. And if you make a habit out of memorizing useful things, this might not be a bad one. – davidlowryduda Aug 28 '11 at 03:05
  • 3
    @Jordan Clayton: Calculate the product $(a-b)(a+b)$. You will get $a^2-b^2$. Yes you need to memorize it, indeed internalize it, so that when you see $a^2-b^2$, you simultaneously see $(a-b)(a+b)$, and vice-versa. – André Nicolas Aug 28 '11 at 03:13
  • I am getting 4/160 –  Aug 28 '11 at 03:20
  • @mixedmath I believe you mean $x(4 - \sqrt{x})(4 + \sqrt{x})$ at the end of your hint. – Austin Mohr Aug 28 '11 at 03:28
  • I keep getting 4/160 –  Aug 28 '11 at 03:30
  • @Jordan Clayton: One way, we get $\frac{4-\sqrt{x}}{x(4-\sqrt{x})(4+\sqrt{x})}$. Do the cancellation, let $x$ approach $16$, we get $1/128$. Alternately, multiply top and bottom by $4+\sqrt{x}$, cancel $16-x$ from top and bottom, then let $x$ approach $16$. Again we get $1/128$. Make sure you are using parentheses correctly. – André Nicolas Aug 28 '11 at 04:06
  • How did you attain that for the numerator? I always get 16-x. –  Aug 28 '11 at 14:38
  • 3
    @Jordan: If you're not even familiar with $a^2-b^2=(a+b)(a-b)$, maybe it would be a good idea for you to work a little on your algebra before trying to do limits. Also from your other questions here, it seems that it's not really the limits that are causing you trouble, but the elementary simplifications. – Hans Lundmark Aug 28 '11 at 15:14
  • I don't really have time for that, I would just fall further behind. I think if I do all these I can get a good review as I recall them after I see them. –  Aug 28 '11 at 15:15
1

Multiply top and bottom by $4+\sqrt{x}$ as you have done. Then factorise the denominator (there's a factor of x) and notice that something cancels. Then the limit should be easy to calculate.

Billy
  • 5,224
  • I did do that, don't see anything that cancels. –  Aug 28 '11 at 03:01
  • @Jordan: Multiply top and bottom by $4+\sqrt{x}$, and then multiply out the top, but leave the bottom as it is. You should get the fraction $\frac{16-x}{x(16-x)(4+\sqrt{x})}$. The 16-x (which is the bit that was causing you the problem by giving you a denominator of 0) will cancel. – Billy Aug 28 '11 at 03:59
  • Oh I guess I did it wrong, on the bottom I got (4-x)(4+x) instead of the 16-x, is that wrong? I am not sure why it would be wrong but it gives the wrong answer. –  Aug 28 '11 at 14:29
  • I don't understand what I am doing wrong. To me this problem is just some magical chunk of numbers that is only able to be solved if you accidently find the correct and arbitrary order of solving it. There are many mathematically correct ways to break down this problem, but only one gives you a correct answer. $\frac{16-x}{(4+x)(4-x)(4+\sqrt{x})}$ –  Aug 28 '11 at 15:07
  • 2
    @Jordan: Maths does not know what you want. There are tons of legitimate things you can do to any problem, and only some will give you the answer you want. Which ones? Well, in most cases that's not easy to answer (that's why mathematical researchers exist!). In this case, though, it's not too bad. Your original problem was obvious: as $x \to 16$, the denominator was heading to zero, and you can't divide by zero. Your task was to find out what factor was going to 0, and (if possible) get rid of it, probably by cancelling with something appropriate in the numerator. That thing was $16 - x$. – Billy Aug 30 '11 at 19:43
  • Incidentally, your factorisation is wrong. But the idea is fine. Factorising $16-x$ (correctly) as $(4-\sqrt{x})(4+\sqrt{x})$ is not "wrong" (of course it's not!). But is it helpful? Actually, yes: now the factor causing you the trouble is $4-\sqrt{x}$, and you can cancel that immediately. On the other hand, factorising it as $(2+\sqrt[4]{x})(8 - 4\sqrt[4]{x} + 2\sqrt{x} - \sqrt[4]{x}^3)$ would have been "correct" but not at all useful. – Billy Aug 30 '11 at 19:47
1

Jordan,

I went through all of your limit problems,computed them but couldn't post the solution as it was already there. Frankly i found them very usefull as i'am a beginner too. Here's the solution for the last one.

$\lim\limits_{x \to 16}\; \dfrac{(4-\sqrt x)}{(16x-x^2)}$

$(16x - x^2) = x(16 - x) = x(4 - \sqrt x)(4 + \sqrt x)$

Use the above hint and you'll get lim x-->16 (16-x)/x(16-x) = 1/16.I'am sorry for not using that crap called Latex or whatever,I'am realy having loads of trouble with it and it's very annoying.Hope i get somehelp in formatting math expressions.

Hope you found that helpfull.Remember continue to ask as many as questions when you are in doubt.This website is purely dedicated for math.Never think that you are wasting someone else's time.If they don't want to answer chuck them.If someone says that your questions are stupid just ignore them.[fortunately no one has ever said so far].Here are a bunck of questions that i had previously asked on Limits that should be of great help.

limits and Continuity

One last question on the concept of limits

Good Luck Hope to see you soon here.

Community
  • 1
alok
  • 3,890
  • Jordan:If you have further questions related to your homework please post them.I'll try to answer them.Infact you can improve your algebra skills along the way. – alok Sep 01 '11 at 10:57
  • Given the way you feel about $\LaTeX$, I now feel sorry for inflicting it on you in your other question. – Srivatsan Sep 01 '11 at 11:12
  • It's ok.It's annoying for every new user,but as they pick it up it becomes interesting. – alok Sep 01 '11 at 13:31
0

Try to use the change of variable. You can see the entire work here. Change of variable

Vinicius L. Beserra
  • 527
  • 1
  • 5
  • 14