Finding function limit LIM (Sinn) / n Finding function limit LIM (n!) / N ^ n N tends to infinity Why?
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- 1. Find function limit LIM (n-1) * 2 ^ n, n tends to be positive infinity
- 2. Find the limit LIM (x → 1) (x - N - 1) / (x-1) =? N is a positive integer
- 3. Lim X - > 0 + X / (1-cosx) ^ 1 / 2 why is the answer 2
- 4. X →∞, Lim [(x ^ 2 / 1 + x) + ax-b] = 1, find a, B
- 5. Proof of mathematical limit: LIM (n-positive infinity) [(- 1) ^ n / N ^ 2] = 0
- 6. Lim √ (n & # 178; - A & # 178;) / N = 1 proved by definition N approaches positive infinity
- 7. Prove LIM (n →∞) (1 + 1 / N & # 178;) = 1 by definition
- 8. Lim((1²+2²+…… +n²)/n³)n→∞=
- 9. Please help me prove LIM (1 + 1 / N) ^ n = e, thank you!
- 10. lim n→∞ [(n!)^1/n]/n = 1/e Our advanced mathematics teacher uses the idea of LIM an = a, Lim BN = (a1 + A2 +... + an) / N = a
- 11. lim n->0 (1+3n/1)^3n = ? lim n->0 (1+3n/1)^3n = ?
- 12. Fill in the blanks LIM (n →∞) = √ (3N ^ 2 + 6N + 5) / (3n-2)=____ How to calculate this problem?
- 13. limn→∞[11•4+14•7+17•10+… +1(3n−2)(3n+1)]=______ .
- 14. lim(1+4+7+… +(3n-2))/(7n^2-5n),
- 15. Find the following limit, Lim e ^ (1 / 2x), X → 0 Please give the specific steps,
- 16. The calculation limit is 2 / X of LIM (n →∞) (1 + 2x) (x→∞)
- 17. LIM (arctan x-x) / SiNx ^ 3 x tends to 0, can we use the substitution of equivalent infinitesimal to solve The answer is - 1 / 3, according to the law of lobita In addition, I use the Equivalent Infinitesimal Substitution, the result is - 1 / 6 =(sin x-x) / SiN x ^ 3 x tends to zero =(cosx-1) / [(cosx ^ 3) * 3x ^ 2] x tends to 0 =(- 1 / 2) x ^ 2 / [(cosx ^ 3) * 3x ^ 2] x tends to 0 =(- 1 / 2) / 3cosx ^ 3 x tends to zero =-1/6
- 18. lim(x→0)((arctan(x∧2))╱(sin(x╱2)sinx))
- 19. Can x → 0 Lim arctan (SiNx / x) be written as arctan LIM (SiNx / x)? Can Lim ln (SiNx / x) be written in this form? Why?
- 20. Find Lim e ^ (1 / x ^ 2) * arctan ((x ^ 2 + 2-1) / (x + 1) * (X-2)) when x approaches 0, - 1,2