Determining Limits Algebraically

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How to find the derivative of sinx using limits (version II)

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Key Questions

  • When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

    #lim_{x to 0^-}1/x=1/{0^-}=-infty#

    1 is divided by a number approaching 0, so the magnitude of the quotient gets larger and larger, which can be represented by #infty#. When a positive number is divided by a negative number, the resulting number must be negative. Hence, then limit above is #-infty#.

    (Caution: When you have infinite limits, those limts do not exist.)

    Here is another similar example.

    #lim_{x to -3^+}{2x+1}/{x+3}={2(-3)+1}/{(-3^+)+3}={-5}/{0^+}=-infty#

    If no quantity is approaching zero, then you can just evaluate like a two-sided limit.

    #lim_{x to 1^-}{1-2x}/{(x+1)^2}={1-2(1)}/{(1+1)^2}=-1/4#

    I hope that this was helpful.

  • By eliminating common factors, we can find
    #lim_{a to 0}{(a-3)^2-9}/a=-6#.

    Let us look at some details.

    First, we notice that both the numerator and the denominator approach #0# as #a# approaches #0#, which indicates that they share a common factor #a#.

    #lim_{a to 0}{(a-3)^2-9}/a#

    by multiplying out the square,

    #=lim_{a to 0}{a^2-6a+9-9}/a#

    by cancelling out the 9's,

    #=lim_{a to 0}{a^2-6a}/a#

    by factoring out #a#,

    #=lim_{a to 0}{a(a-6)}/a#

    by cancelling #a#'s,

    #=lim_{a to 0}(a-6)#

    by plugging in #a=0#,

    #=0-6=-6#

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