Solution Lim X Sin X 0 Sin 0 First Note That We Cannot Rewrite The Limit As The Product Of Limits Lim X4 And L 1 (43.03 KiB) Viewed 12 times
SOLUTION lim x sin x → 0 ≤ sin - < = 0 First note that we cannot rewrite the limit as the product of limits lim x4 and lim sin x → 0 x → 0 approaches 0 of sin(1/x) does not exist (see this example). 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) = xª 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. = because the limit as x Any inequality remains true when multiplied by a positive number. We know that x4 ≥ 0 for all x and so, multiplying each side of inequalities by x4, we get ≤ x² sin( =) = [ as illustrated by the figure below. We know the following. lim x4 and lim (-x) = x → 0 X→ 0
≤ sin Any inequality remains true when multiplied by a positive number. We know that x4> 0 for all x and so, multiplying each side of inequalities by x4, we get lim x → 0 x4 < as illustrated by the figure below. We know the following. and ximo (−+4) = = 4 sir (2) s (²) < x4 sin X → 0 Taking f(x) = −x4, g(x) = x4 sin(1/x), and h(x) = x 4 in the Squeeze Theorem, we obtain the following. x* - 0 (²) - lim x + sinl
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!