Show that the following limit is true.()-lim x* sin= 0SOLUTIONFirst note that we cannot rewrite the limit as the product of limits lim x4 and lim sinX → 01because the limit as x approaches 0 of sin(1/x) does not exist (see this example).x* sin(1/x) and a function h bigger than g such that bothwe can write the following.Instead, we can find the limit by using the Squeeze Theorem. To apply the Squeeze Theorem, we need to find a function f smaller than g(x)f(x) and h(x) approach 0 and x → 0. To do this we use our knowledge of the sine function. Because the sine of any number lies betweenand< sinAny inequality remains true when multiplied by a positive number. We know that x* 2 0 for all x and so, multiplying each side of inequalities by x*,we get1sx* sinas illustrated by the figure below. We know the following..4%3Dlim (-x*) =limandTaking f(x) = -x*, g(x) = x* sin(1/x), and h(x) = x* in the Squeeze Theorem, we obtain the following.lim x* sin1= 0y

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Show that the following limit is true. lim x sin = 0 SOLUTION First note that we cannot rewrite the limit as the product of limits lim x4 and lim sin because the limit as x approaches 0 of sin(1/x) does not exist (see this example). x- 0 Instead, we can find the limit by using the Squeeze Theorem. To apply the Squeeze Theorem, we need to find a function f smaller than g(x) = x sin(1/x) and a function h bigger than g such that both f(x) and h(x) approach 0 and x - 0. To do this we use our knowledge of the sine function. Because the sine of any number lies between and , we can write the following. s sin Any inequality remains true when multiplied by a positive number. We know that x* 2 0 for all x and so, multiplying each side of inequalities by x*, we get sx* sin as illustrated by the figure below. We know the following. lim and lim (-x*) = Taking f(x) = -x*, g(x) = x* sin(1/x), and h(x) = x* in the Squeeze Theorem, we obtain the following. lim x sin = 0

Show that the following limit is true. lim x sin = 0 SOLUTION First note that we cannot rewrite the limit as the product of limits lim x4 and lim sin because the limit as x approaches 0 of sin(1/x) does not exist (see this example). x- 0 Instead, we can find the limit by using the Squeeze Theorem. To apply the Squeeze Theorem, we need to find a function f smaller than g(x) = x sin(1/x) and a function h bigger than g such that both f(x) and h(x) approach 0 and x - 0. To do this we use our knowledge of the sine function. Because the sine of any number lies between and , we can write the following. s sin Any inequality remains true when multiplied by a positive number. We know that x* 2 0 for all x and so, multiplying each side of inequalities by x, we get sx sin as illustrated by the figure below. We know the following. lim and lim (-x) = Taking f(x) = -x, g(x) = x* sin(1/x), and h(x) = x* in the Squeeze Theorem, we obtain the following. lim x sin = 0

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.6: Permutations

Problem 47E

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