ALA Show that the following limit is true. 1 lim 8 sin x 0 x First note that we cannot use 1 lim 8 sin = lim x 0 I x 0 sin 1 because the limit as a approaches 0 of sin does not exist (see this example). Instead we apply I 1 x 1 x the Squeeze Theorem, and so we need to find a function f smaller than g(x) = x sin function h bigger than g such that both f(x) and h(x) approach 0. To do this we use our knowledge of the sine function. Because the sine of any number lies between and we can write < lim sin x 0 Taking f(x) = x, g(x) 1 lim sin = 0. x 0 I 1 X Any inequality remains true when multiplied by a positive number. We know that 0 for all x and so, multiplying each side of inequalities of x8, we get < sin S as illustrated by the figure. We know that lim 8 2-0 and lim 2 0 and a = = x8 sin -1/4 , and h(x) = 8 in the Squeeze Theorem, we obtain