# Re: very quick help with limits

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Posted by nick on August 28, 2002 at 20:27:29:

In Reply to: Re: very quick help with limits posted by Soroban on August 28, 2002 at 17:31:50:

: Nick,

: Ryan's advice is correct: L'Hopital is the way to go.

: But from the nature of the problems, though, I suspect that L'Hopital
: is NOT the method they expect. In fact, the problems look like they
: come from a section *before* L'Hopital is taught.
: Can we find the limits anyway? Yes!

: Two of them require knowledge of this theorem:
: lim (sin z)/z = 1 (as z -> 0)

: You probably saw it first when differentiating the sine function -
: and thought (or hoped) you'd never see it again. *LOL*
: You'll be surprised at how it can be used. (I remember that *I* was.)

: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

: : lim (sin x)/(2x²-x) [x -> 0]

: Factor the denominator and write the expression as: (sin x)/x(2x - 1)

: Then write it as two fractions: [(sin x)/x](1/(2x-1)]

: Take the limit [x -> 0]:
: lim (sin x)/x * lim[1/(2x-1)]

: The first limit is 1, the second is -1:

: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

: : lim (3sin 4x)/(sin 3x) [x -> 0]

: This one takes some acrobatics.
: we have: (3 sin 4x)/(sin 3x)

: Divide numerator and denominator by 3x:
: [(sin 4x)/x]/[(sin 3x)/3x]

: In the numerator, multiply by 4/4:
: [4(sin 4x)/4x]/[(sin3x)/3x]

: Now, we take the limit [x -> 0]:
: 4 * lim[(sin 4x)/4x] / lim[(sin 3x)/3x]

: Both limits equal 1. Answer: 4*1/1 = 4

: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

: : lim (1/(2+x)-1/2)/x [x -> 0]

: This one requires only algebra.
: Simplify the numerator (get a common denominator and combine):
: 1/(2+x) - 1/2 = -x/[2(2+x)]

: Write it over x and cancel:
: -x/[2(2+x)]/x = -1/[2(2+x)

: The limit [x -> 0] is -1/4

: ~~~~~~~~~~~~~~~~~~~~~~~~~~

: We have to be careful when using that theorem:
: lim (sin z)/z = 1 [ z -> 0]

: Note that the "z" in the sine function is
: the SAME as the "z" in the denominator.
: You're thinking "Well, duh!", but bear with me.

: If we have: (sin 2x)/x, the "z" is NOT the same.
: We need a "2x" in the denominator.

: So, we multiply by 2/2: 2(sin 2x)/2x

: And NOW we can take the limit: (sin 2x)/2x -> 1,
: so the limit is 2(1) = 2.

: Got it? It's a tricky thing to digest.
: And it should explain the "acrobatics" in the second problem.

thank you so freakin much. i kinda misunderstood when these problems were do. we didn't learn the (sin x)/x=1 thing until today. i should've gotten tthe algebra one. thank you.

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