Re: very quick help with limits

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: : 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: : : Answer: (1)(-1) = -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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