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Finding limits by factoring (cubic)

Sal finds the limit of (x³-1)/(x²-1) at x=1 by factoring and simplifying the expression. Created by Sal Khan.

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  • blobby green style avatar for user 7imon7ays
    Where can I watch a tutorial on the kind of division Sal does at ?
    (104 votes)
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  • aqualine ultimate style avatar for user Anoir Trabelsi
    is x^3 - y^3 = (x-y)(x² + xy + y²) ?
    (15 votes)
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  • leaf green style avatar for user shahin.imtiaz
    At , Sal writes when x =/= 1. But after simplification, even x=-1 will give an undefined answer. Why do we not take that into consideration? How do we know what to put conditions for and what not to put conditions for.
    I understand that since the question deals with limit of x going to 1, -1 is not that important, but my question is still how do we generally know what to put conditions for.
    (10 votes)
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  • blobby green style avatar for user christylindh
    how do I do the problem:
    Find the limit:
    lim as x approaches 0: {1/(3+x)}-(1/3)/x

    fractions in the numerator...I forget this algebra/simplifying
    (5 votes)
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    • male robot hal style avatar for user Alex Telon
      The problem you have is that you don't want to divide by 0, right? So lets try to get rid of the x'es in the numerator, because if we do that we wont have any problem with letting x go to 0 am I right?
      So my suggestion would be multiplying both denominator and numerator with x and see what happens! :)
      (7 votes)
  • leafers ultimate style avatar for user Abhay Mehta
    I read a popular anomaly in maths where the end result is 2=1 when we take A=B.
    It goes like this- Let A=B
    => A^2 =AB => A^2 - B^2 = AB - B^2 => (A+B)(A-B) = B (A-B) ------ Step n
    => A+B=B
    => 2B=B => 2=1

    The explanation given there was that in step n we divided both sides by zero. As A-B=0, therefore this anomaly came up.

    Here too aren't we doing the same thing?
    (5 votes)
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    • blobby green style avatar for user Creeksider
      Good question. At , Sal is careful to point out that the division he is performing is valid only if we assume x does not equal 1. We make that assumption when performing a limit calculation because we aren't trying to find the value when x = 1. Instead, we're trying to find the value that is approached as x gets closer and closer to 1. As long as we allow for some difference between x and 1, we aren't dividing by zero and the operation is valid.

      Now it may seem like a contradiction that Sal goes on after that to compute the limit by plugging 1 into the resulting fraction, but again you have to keep in mind that he isn't calculating the value of the fraction at 1, he's calculating the value of the limit at 1, which means finding the value that the surrounding points gravitate towards. Plugging in 1 at this point gives us that value.
      (7 votes)
  • leaf green style avatar for user gman9732
    Why was he allowed to use 1 if he said x cannot equal one?
    (4 votes)
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    • leaf red style avatar for user Noble Mushtak
      He was allowed to substitute 1 for x even though there was a restriction of "x does not equal 1" because he was finding the limit of the original expression as x approaches 1. To find the limit, he can substitute 1 for x into the new expression for which x=1 is defined to get the limit for the old expression for which x=1 is not defined. However, this is not the value of the old expression at x=1: It is the limit. That's why he can substitute 1 for x into the new expression despite the restriction.
      (8 votes)
  • male robot hal style avatar for user zollen.kong
    Would anyone able to show me why the following limit is 9? I would be much appreciate if steps are also shown. Thanks!!
    Lim (sin ^2 (3x)) / x^2
    x -> 0
    (3 votes)
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    • leaf grey style avatar for user Qeeko
      (In what follows, it is assumed that x → 0 in every limit statement, if not otherwise stated.)

      One way to prove the identity is by making use of the following preliminary relations:
      1. lim ( sin(x) / x ) = 1;
      2. If lim ƒ(x) = F and lim g(x) = G, both as x → a, then lim ƒ(x)g(x) = FG as x → a, where a is any real number.

      The fact that lim ( sin²(3x) / x² ) = 9 may now be deduced by rewriting sin²(3x) / x² to a form we recognise.

      I) Properties 1. and 2. imply that lim ( sin²(x) / x² ) = 1. To see this, observe that lim sin²(x) / x² = lim ( sin(x) / x )² = lim ( sin(x) / x ) · ( sin(x) / x ). By properties 1. and 2., we have, then, lim sin²(x) / x² = (lim sin(x) / x) · (lim sin(x) / x) = 1 · 1 = 1.

      II) We now compute lim sin²(3x) / (3x)². Let u = 3x. Then u → 0 as x → 0; hence lim sin²(3x) / (3x)² = lim sin²(u) / u² = 1, which follows from I).

      III) Observe that sin²(3x) / x² = sin²(3x) / ([1/9] · 9x²) = 9 · ( sin²(3x) / (3x)² ). By properties II) and 2., it follows that lim sin²(3x) / x² exists and equals (lim 9) · (lim sin²(3x) / (3x)²) = 9 · 1 = 9.

      This completes the proof.
      (5 votes)
  • aqualine ultimate style avatar for user Nikhil Naidu
    If we differentiate the numerator and denominator by L'hospital rule that would give the same answer right?
    (3 votes)
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    • piceratops ultimate style avatar for user Just Keith
      If you are referring to the problem in the video, that is a 0/0 form and thus is valid for l'Hopital's. So, yes, you could do it that way (and I would definitely solve it that way)

      lim x→1 (x³-1)/(x²-1)
      = lim x→1 (3x²)/(2x)
      = ³⁄₂
      However, you should not discount the techniques Sal used in the video because you cannot use l'Hopital's on many difficult limits, so it is good to know the techniques for finding the limits without l'Hopital's.

      Furthermore, even when you can use l'Hopital's, it is not always the easiest method.
      (6 votes)
  • piceratops ultimate style avatar for user ASDF ASDF
    at how does x^2-1 becomes (x-1)(x+1)?
    (1 vote)
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  • piceratops ultimate style avatar for user Ali Ahmad
    Isn't (x-1)/(x-1) = 0/0 ? Are you not canceling 0 with 0?
    (1 vote)
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Video transcript

Let's try to find the limit as x approaches 1 of x to the third minus 1 over x squared minus 1. And at first when you just try to substitute x equals 1, you get 0/0 1 minus 1 over 1 minus 1. So that doesn't help us. So let's see if we can try to simplify this in some way. So you might immediately recognize-- so let's rewrite this expression right over here so it's x to the third minus 1 over x squared minus 1. This on the bottom immediately jumps out as a difference of squares. So we know on the bottom that this could be factored as x minus 1 times x plus 1. And so if somehow this thing on the top also has an x minus 1 as a factor, then that x minus 1 will cancel with this, and then we're not going to have an issue of dividing by 0. The reason why I care about the x minus 1 term is that this is what's making our denominator equal 0. When you say x equals 1, you have 1 minus 1 times 1 plus 1. So 0 times 2, it's this 0 that's making our denominator 0. So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x goes into x to the third x squared times. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing. We can have these cancel out if we assume x does not equal 1. So that is equal to x squared plus x plus 1 over x plus 1, for x does not equal 1. And that's completely fine, because we're not evaluating x equals 1. We're evaluating as x approaches 1. So this is going to be the same thing as the limit as x approaches 1 of x squared plus x plus 1 over x plus 1. And now this is much easier to find. You could literally just say, well, what happens as we get right to x equals 1? Then you have 1 squared, which is 1 plus 1 plus 1, which is 3, over 1 plus 1, which is 2. So we get that equaling 3/2.