Question

高数。求解答过程，谢谢。

Answer 1

令t=1/x，则t→0
原式=lim[tant/t]^(1/t²)（）
=lim[1+(tant-t)/t]^[t/(tant-t) · (tant-t)/t^3]
=lime^[(tant-t)/t^3]
=lime^[(sec²t-1)/3t²]
=lime^[(1-cos²t)/3t²cos²t]
=lime^[(1-cost)(1+cost) / 3t²cos²t]
=lime^[½t²(1+1)/3t²·1]
=e^(1/3)

Answer 2

[(tanx)/x]^(1/x²)
= e^ln[(tanx)/x]/x²
= e^[ln(tanx) - lnx]/x²
lim(x→0) [ln(tanx) - lnx]/x²,0/0型,洛必达法则
= lim(x→0) (sec²x/tanx - 1/x)/(2x)
= lim(x→0) [1/(sinxcosx) - 1/x]/(2x)
= lim(x→0) (x - sinxcosx)/(2x²sinxcosx)
= lim(x→0) [x - (1/2)sin2x]/(x²sin2x),0/0型,洛必达法则
= lim(x→0) (1 - cos2x)/(2x²cos2x + 2xsin2x)
= lim(x→0) [1 - (1 - 2sin²x)]/(2x²cos2x + 2xsin2x)
= lim(x→0) 2sin²x/(2x²cos2x + 2xsin2x)
= lim(x→0) x²/(x²cos2x + xsin2x),sin²x x²当x→0
= lim(x→0) x/(xcos2x + sin2x)
= lim(x→0) 1/[(xcos2x + sin2x)/x]
= lim(x→0) 1/[cos2x + (sin2x)/(2x) · 2]
= 1/(1 + 2)
= 1/3
∴lim(x→0) [(tanx)/x]^(1/x²) = e^(1/3)

Answer 3

呼

Answer 4

如图所示，满意请采纳