9this is a very gud Q involving limit of limits !!!!
i m stating some things u try finding reasons urself (simple only)
lt exists at 1
means lt at 1- = lt at 1 = lt at 1+
now lt at 1 = f(1) +g(1) /2
lt at 1- = g(1-)
lt at 1+ = f(1+)
now all these equal imply
f(1) = g(1)
( note its already gvn f,g are difff so theyr continuous and f(1+)=f(1)=f(1+) )
so we know for sure f(1) =g(1)
hence at least 1 root for f(x)=g(x) exist btw 0,2