Lim sin (x-1) / X-1 x tends to infinity

RELATED INFORMATIONS

• 1. How much is SiNx, sin (1 / x), sin (x ^ 2) equal when Lim x approaches zero? And how much are they equal when Lim x approaches infinity? And how much is cos Tan in this case?
• 2. lim(x->0)[(x^2)*sin(1/x)]/sinx I use the equivalent infinitesimal, because when (x - > 0), SiNx ~ x, so sin (1 / x) ~ 1 / x, Then the original formula = LIM (x - > 0) (x ^ 2) * (1 / x) / SiNx = 1, right
• 3. LIM (x tends to infinity) (sin n) / N =?
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• 8. LIM (sin △ X / 2) / △ x is a 1 / 2 proof process when △ x tends to zero
• 9. [mathematical analysis] proves that LIM ∫ (sin (x)) ^ n = 0 The book emphasizes that if the mean value theorem is used, then there should be ξ_ n→π/2 In the end, it shows that for 0
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• 12. Find Lim X - > ∞ (xtan2 / x + (1 / x ^ 2) * (SIN) x ^ 2)
• 13. lim(x---1)[sin^2(x)-sin^2(1)]/[x-1]=? University Advanced Mathematics, seek advice
• 14. Find Lim sin (x-1) / (x ^ 2-1) where X - > 1
• 15. Detailed process of LIM (x → 1) lncos (x-1) / (1-sin (π X / 2))
• 16. Limit: LIM (x - > 4) (x ^ 2-6x + 8) / (x ^ 2-5x + 4)
• 17. Finding limit LIM (x ^ 2-6x + 8) / x ^ 2 + X-6
• 18. 1.5 find the limit LIM (x → 4) (x ^ 2-6x + 8) / (x ^ 2-5x + 4) 1.5 find the limit LIM (x → 4) (x ^ 2-6x + 8) / (x ^ 2-5x + 4)
• 19. Help find a limit: LIM (x ^ 2-6x + 8) / (x ^ 2-5x + 4) x → 4
• 20. lim x→∞ x^2-6x+8/x^2-5x+4 Wrong. It's not infinity. It's four