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Limit help!

  1. May 10, 2005 #1

    Find tangent slope to parabola using Theorem 2.
    [tex]y(x) = x^2 + 2x \; \text{at} \; P(-3.3)[/tex]

    Theorem 2:
    [tex]m = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}[/tex]
    [tex]m = \lim_{h \rightarrow 0} \frac{(a + h)^2 + 2(a + h) - 3}{h} = \lim_{h \rightarrow 0} \frac{a^2 + h^2 + 2ah + 2a + 2h - 3}{h}[/tex]
    [tex]\lim_{h \rightarrow 0} \frac{a^2 + h^2 + 2ah + 2a + 2h - 3}{h} = \lim_{h \rightarrow 0} \frac{(a + h - 1)(a + h + 3)}{h}[/tex]
    :uhh:

    I have already solved the tangent line using the Tangent Line Theorem, however, I have been unable to eliminate [tex]h[/tex] from the denominator in this theorem using division or numerator conjugates...

    Any suggestions?
     
    Last edited: May 10, 2005
  2. jcsd
  3. May 10, 2005 #2

    mathman

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    Gold Member

    either use a throughout or immediatedly plug in its value. Then
    m=lim (h2+2ah+2h)/h=2(a+1)
     
  4. May 10, 2005 #3

    dextercioby

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    Isn't a=-3...?

    Daniel.
     
  5. May 10, 2005 #4
    [tex] \lim_{h\rightarrow 0} \frac{y(x+h) - y(x)}{h} [/tex]

    y(x) = x^2 + 2x

    [tex] \lim_{h\rightarrow 0} \frac{(x+h)^2 + 2(x+h) - x^2 - 2x}{h} [/tex]

    [tex] \lim_{h\rightarrow 0} \frac{x^2 + 2xh + h^2 + 2x + 2h - x^2 - 2x}{h} [/tex]

    [tex] \lim_h{\rightarrow 0} \frac{2xh + h^2 + 2h}{h} [/tex]

    Can you see whats going on now?
     
  6. May 10, 2005 #5

    Solved!, thanks everyone!
     
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