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# squeeze theorem?

i know the answer is 4/3 but i don't know how to get there

### 2 Answers

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- atsuoLv 65 months agoFavourite answer
Let u = 1/x. So

[lim x→+inf] (4x^3 - 1)/(3x^2 + 2) * sin(1/x)

= [lim u→+0] (4/u^3 - 1)/(3/u^2 + 2) * sin(u)

= [lim u→+0] [(4/u^3 - 1)u^3]/[(3/u^2 + 2)u^2] * sin(u)/u

= [lim u→+0] [4 - u^3]/[3 + 2u^2] * sin(u)/u

= 4/3

note. We know [lim u→+0] sin(u)/u = 1

- 5 months ago
Basic circle inequality sin(y)<y<tan(y) is true for all small y>0

Rewrite it as ycos(y)<sin(y)<y

(4x^3-1)/(3x^2+2) (1/x)cos(1/x) < (4x^3-1)/(3x^2+2) sin(1/x)< (4x^3-1)/(3x^2+2) (1/x)

(4x^3-1)/(3x^3+2x) cos(1/x) < (4x^3-1)/(3x^2+2) sin(1/x) < (4x^3-1)/(3x^3+2x)

As x->inf (4x^3-1)/(3x^3+2x)->4/3 and cos(1/x)->1

Hence 4/3 .le. (4x^3-1)/(3x^2+2) sin(1/x) .le. 4/3

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