27 thoughts on “Nice aswers

  1. I beg to differ with the first example, unless we’re talking about a level of mathematics I have not yet reached, the limit as x approaches 8 does not exist. as it approaches from the left the limit is negative infinity and as it approaches from the right it is positive infinity, making 8 a verticle asymtote with an indeterminate limit. if im wrong(such that maybe the taylor series determines the limit), please correct me curiousity astounds me as to how the first example is correct.

  2. It is just notation. Limit can be a positive infinity, a negative infinity or infinity. The latter means a) that absolute value, |f(x)| has positive infinity as its limit; b) that reciprocal, 1/f(x), has zero as its limit; c) that the expression has infinity as its limit when considered as a mapping with image on the circle which goes from the only infinity through negative numbers, zero, and positive numbers back to infinity (consider a line l with numbers on it an a point X outside; consider all (straight) lines through X; you’ll have a line for each number on l, and one more – for single infinity; incline it left or right and it will become a big negative or a big positive number).

  3. David,

    As x approaches 8, the denominator approaches zero, yielding a limit of infinity for the expression in question.

    The Taylor series is a representation or approximation of a function as a sum of terms calculated from the values of its derivatives at a single point. It has nothing to do with the example.

  4. I agree with the first comment. Unless there was a typo and a ‘+’ sign was suppose to be place right next to x -> 8, the limit does not equal to infinity.

  5. Actually, David is quite correct in stating that the limit does not exist. The problem with understanding this problem is that we DO have to consider taking the limit from both the left AND the right hand sides. In this case, take a look at what happens when x is 7.9 – close to 8, but not quite. The function evaluated at x=7.9 is:
    -10
    when x is 7.999 – closer to 8, but not quite. Then, the function is equal to:
    -1000
    when x is 7.99999999 – closer to 8 – the function results in:
    -100000000
    and so on – you can see it approaches -Infinity – from the Left

    Now, from the right:
    when x = 8.1, then f(x) = +10
    when x = 8.0001, then f(x) = +10000

    and so on – you can see it approaches +Infinity – from the Right.

    While you’re right about it not having anything to do with Taylor series, David is right on the mark about the function not existing as x approaches 8.

  6. As 8 moves closer to 8 (in numerical terms) 7.88, 7.98, 7.999, etc…The closer to 0 the denominator becomes. What does 1/0 equal? Or, how many times can you divide 0 into1? You can’t. So as the denominator goes closer to 0 (ex. 1/.0001) the larger the value. So the closer x gets to 8, in the denominator, the more infinately large the number becomes.

    ex. 1/.1=10, 1/.001= 100, 1/.0001= 1000, 1/.000001=10,000, etc…..

  7. Angel,

    As x approaches 8 from the right – that is, numbers greater than 8, but progressively smaller and smaller, approaching 8 – then the quotient begins to look like 1 divided by an incredibly small positive number, and thus as the denominator gets smaller and smaller (but still positive) then the quotient blows up to infinity. The limit as x approaches 8 from the right is infinity.

    As x approaches 8 from the left however – meaning that values used are smaller than 8 but get progressively larger and larger, approaching 8 – then the quotient begins to look like 1 divided by an incredibly small negative number (because x

  8. Sorry, wasn’t aware there was a character limit…

    … begins to look like 1 divided by an incredibly small negative number (because x

  9. David is correct. There is NO limit. The inventor of the problem looks more foolish than the student who answered. A limit of infinity only exists if all n>0, there exists a d>0, such that for all x, |x-8| n.
    Choose some n and let d’>0 be the corresponding value of d satisfying the above property. Choose x = 8 – (d’/2) which clearly satisfies |x-8|0 as required by the property.

    Hence lim x->8 (1/(x-8)) cannot exist.

  10. Oh great. This web form submit doesnt accept the less than sign and screws up the explanation. So rewriting it …

    A limit of infinity only exists if all n>0, there exists a d>0, such that for all x, |x-8| is less than d implies that 1 / (x-8) > n.
    Choose some n and let d’>0 be the corresponding value of d satisfying the above property. Choose x = 8 – (d’/2) which clearly satisfies |x-8| less than d. Then 1/(8-(d’/2)-8) = -2/d’. This value -2/d’ can never be positive and therefore can never be greater than n as required by the property.

    Hence lim x->8 [1/(x-8)] cannot exist.

  11. I’m afraid that I think David is correct. 1/(7-8) equals -1. 1/(9-8) equals 1. Since the left hand limit and the right hand limit do not equal each other, the limit does not exist.

  12. Actually, if you take it as the limit of 1/(x-8) as x approaches 8 form the positive, it is negative infinity. From the negative it’s positive infinity. But it’s a friggin JOKE, guys, arguing semantics will not make it any funnier.

  13. it approaches negative infinity. 7.9999999999-8= -.000000001 and so on. So if it doesn’t matter if it’s positive or negative infinity, then the equation is correct, otherwise it’s missing a negative sign in front of infinity

  14. well, if something that is aproaching is ininit, then it also is starting from what ever it is aproaching, meaning its already there, no matter if its coming from the left, right, back, front, top, or bottum. and last i was told infinit means for ever going, so how can you get a number out of something that doesnt stop, and you can not messure with any numbers because you would just keep adding more and more numbers to the never ending thing. so if what ever it is you are trying to solve would be. if it deals with anything infinit to help solve it. they answer would always remain a “X” or “Y” or what ever letter is being put into the problem =D

    P.S. dont know math, its randomness, and my logic. math doesnt always work with logic, but it still seams to work out somehow =D

    oh ya, and i hope someone tries to explain it, for people who see this in the future and dont know what the hell is going on… like me ^^

  15. You could say the limit doesn’t exist because the left and right side limits are unequal. You could say the limit does exist if you view the complex plane as the Riemann Sphere in which there is only one point at infinity, in this case infinity = -infinity. Depending on your view. As a beginner, you should think of the limit not existing, for the first reason and the second is because the limit is unbounded.

  16. No you fools, Angel is correct. Observe:

    x=1
    f(x)=-1/7

    x=9
    f(x)=1

    x=8.1

    f(x)=10

    x=7.9999
    f(x)=10,000.0000000233
    and so on

  17. Yes David is definitely correct. You could use the response “infinity” to describe a quantity that increases without upper bound. So for example, as x approaches 8 from the right ( meaning values of x > 8 ), you could describe the limit by saying it “approaches infinity.” But definitely this is different from what happens as x approaches 8 from the left ( values of x

  18. OK then… just relax and imagine a little “+” sign by the “lim”… does that make you happy? What geeks, picking apart a math joke… I mean, let me say this again: you’re taking a math joke seriously. Just think about that.

  19. Okay if the limit approaches 8 from the right, then it would be infinity. sheesh

  20. @People saying “Stop taking this so seriously”: Consider what would happen if you replaced the ‘blonde’ in any blonde joke with a brunette. People would cry that the joke makes no sense. Someone from the outside who doesn’t understand the stereotype that blondes are dumb would exclaim, “Stop taking a joke so seriously, it screwed up a stereotype, sheesh”. That sounds ridiculous, but that’s exactly how you guys sound.

  21. @Mike: You can type the less-than sign by typing an ampersand, then an lt, and then a semicolon, I believe. Example: < (If that doesn’t convert, forgive me…)

    The less-than sign is only rejected because it’s also an angled bracket, which is used to mark up HTML.

  22. Wouldn’t the joke be just as funny if it used a different function containing 8?

  23. That limit does not exist. First joke is an idiocy, full stop. Especially with the introduction given, the only think making me laugh is “what an idiot teacher”. Let me put it like that:
    “After explaining to a biology student through various lessons and examples that every horse has five legs, I tried to check if she really understood that, so I gave her a different animal. This was the result: The dog has three legs.”
    Funny isn’t it?
    Idiots.

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