SOLUTION First note that we cannot use because the limit as x approaches 0 of si
ID: 2871102 • Letter: S
Question
SOLUTION First note that we cannot use because the limit as x approaches 0 of sin(1/x) does not exist (see this example). Instead we apply the Squeeze Theorem, and so we need to find a function f smaller than g(x) = x^2 sin(1/x) and a 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 Any inequality remains true when multiplied by a positive number. We know that x^2 > = 0 for all x and so, multiplying each site of inequalities by x^2 , we get as illustrated by the figure. We know that Taking fx) = -x^2, g(x) - x^2 sin(1/x), and h(x) = x^2 in the Squeeze Theorem, we obtainExplanation / Answer
-1
1
-1 <= sin(1/x) <= 1
-x^2 <= x^2 * sin(1/x) <= x^2
lim x --> 0 x^2 equals 0^2 = 0
lim x ---> 0 -x^2 equals -0^2 = 0
So, here is what you must enter into the blanks from top to bottom :
-1
1
-1
1
-x^2
x^2
0
0
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