Question

求极限！

Answer 1

原式=lim(x->+∞)e^ln(π/2-arctanx)^(1/lnx)
=e^lim(x->+∞)【ln(π/2-arctanx)】/(lnx)
=e^lim(x->+∞)【1/(π/2-arctanx)·(-1/(1+x²))】/(1/x)
=e^-lim(x->+∞)[x²/x(π/2-arctanx)](1+x²)】
=e^-lim(x->+∞)(1/x)/(π/2-arctanx)]
=e^-lim(x->+∞)(-1/x²)/(-1/(1+x²))]
=e^-lim(x->+∞)((1+x²)/x²)
=e^(-1)

Answer 2

取自然对数得
lim(x→∞）ln（π/2-arctanx)/lnx (这是∞/∞型，运用洛必达法则）
＝lim(x→∞）-1/(1+x^2)*1/（π/2-arctanx)*x
=lim(x→∞）-1/x*1/（π/2-arctanx)(这是0/0型，运用洛必达法则）
=lim(x→∞）1/x^2*[-(1+x^2)]
=-1
其中lim(x→∞）-1/(1+x^2)*x=lim(x→∞）-1/x
所以lim(x→∞）（π/2-arctanx)^(1/lnx)=1/e

Answer 3

答案是e的-1次方哦亲~

Answer 4

syms x
limit((pi/2-atan(x))^(1/log(x)),x,+inf)

ans =

exp(-1)