Problem: If x=1+1+1+1+……x=\sqrt{1+ \sqrt{1+ \sqrt{1+ \sqrt{1+ \ldots \ldots}}}}x=1+1+1+1+……, then:
Answer Choices:
A. x=1x=1x=1
B. 0<x<10<x<10<x<1
C. 1<x<21<x<21<x<2
D. xxx is infinite
E. x>2x>2x>2 but finite Solution:
x=1+x,x2=1+x,x2−x−1=0x=\sqrt{1+x}, x^{2}=1+x, x^{2}-x-1=0x=1+x,x2=1+x,x2−x−1=0, and x∼1.62x \sim 1.62x∼1.62.
∴1<x<2.\therefore 1<x<2 . ∴1<x<2.