Question

求红线处极限，求详细过程

Answer 1

lim(x→∞) (x²+x+1)/[(x-1)(x+2)]
=lim(x→∞) [1+(1/x)+(1/x²)]/[(1-1/x)(1+1/x)].....分子分母同时除以x²
=(1+0+0)/[(1-0)(1+0)]
=1
∴lim(x→∞) arctan(x²+x+1)/[(x-1)(x+2)]
=arctan1
=π/4
lim(x→∞) e^(1/x²)
=e^0
=1
∴lim(x→∞) e^(1/x²)*arctan(x²+x+1)/[(x-1)(x+2)]
=lim(x→∞) e^(1/x²)*lim(x→∞) arctan(x²+x+1)/[(x-1)(x+2)]
=1*(π/4)
=π/4