How prove lim x>0 (1-cosx/x)=0 Solution The squeeze theorem basically says if yo

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How prove lim x>0 (1-cosx/x)=0

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The squeeze theorem basically says if you have two functions that bound another function, and both of them have a particular limit at a particular point, then third function has that same limit at that same point. More formally, if you have three functions f(x), g(x), h(x) that are all defined on some interval I, except possibly at a single point a, and f(x) a h(x), then lim x->a g(x) = L. It's called the squeeze theorem because the functions f(x) and h(x) "squeeze" the function g(x) at the limit point. This is useful, because if you have a function where you can't easily prove the limit directly, but you can find two other functions that have the same limit at the point of interest, and they satisfy the other conditions of the theorem, you get the limit of function you're interested in "for free". In the case of lim x->0 sin(x)/x, g(x) in the theorem would be g(x) = sin(x)/x. For h(x), we can use h(x) = 1 (no point in making it hard on ourselves :)) The function f(x) = 1 - x^2 is less than or equal to sin(x)/x in any interval containing 0. Since f(x) 0 h(x) are both equal to 1, the limit lim x->0 sin(x)/x = 1. In the case of lim x->0 (1-cos x)/x, g(x) = (1 - cos x)/x. This function is very nearly equal to x/2 in a small interval around 0, which bounds it above for x>0, and bounds it below for x0 f(x) = 0, and lim x->0 h(x) = 0, so lim x->0 g(x) = 0

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