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Using limits to find tangents

  1. Apr 21, 2012 #1
    I've been using the formula
    lim
    x-->a f(x)-f(a)
    --------
    x-a
    I haven't had any problems until I was asked to find the slope of the tangent at the general point whose x-coordinate is a. How do I do it with an a instead of a point?

    I'm trying to find it for x^2+x+1

    Thanks!
     
  2. jcsd
  3. Apr 21, 2012 #2


    [itex]\lim_{x\to a}\frac{x^2+x+1-(a^2+a+1)}{x-a}=\lim_{x\to a}\frac{(x-a)(x+a+1)}{x-a}=\lim_{x\to a} (x+a+1)=2a+1[/itex]

    DonAntonio
     
  4. Apr 21, 2012 #3
    I got to the first step, but how do you get to the second step?
     
  5. Apr 22, 2012 #4

    Office_Shredder

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    By the second step do you mean when he factors the numerator, or when he cancels the x-a from the numerator and the denominator?
     
  6. Apr 22, 2012 #5
    [itex]\displaystyle \lim_{x \rightarrow a} \frac{x^2+x+1-(a^2+a+1)}{x-a} = \lim_{x \rightarrow a}\frac{x^2+x-a^2-a}{x-a} = \lim_{x \rightarrow a}\frac{x^2+ax+x-a^2-ax-a}{x-a}[/itex]

    Do you see what to do from there? Creatively adding 0 or multiplying by 1 shows up a lot in mathematics.
     
  7. Apr 22, 2012 #6


    I think that when when we reach some level in mathematics, high school tricks only can, sometimes,

    confuse and/or make things lengthier and more cumbersome.

    Adding and substracting that "ax" in the numerator in one of those tricks, and in this case it just adds more writing innecessarily, imo.

    The best, I think, is to observe that we have a difference of squares in the numerator and thus directly to factor:

    [itex]x^2+x-a^2-a=x^2-a^2+x-a=(x-a)(x+a)+x-a=(x-a)(x+a+1)[/itex] ...

    DonAntonio
     
  8. Apr 22, 2012 #7
    Hmm, I didn't think of it that way! I like that way better!
     
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