1. If the set $A=\left\{\lg x, \lg y, \lg \left(x+\frac{y}{x}\right)\right\}$ has a subset $B=\{0,1\}$, then the maximum possible value of the largest element in $A$ is ( ).
$-、 1 . C$.
If $\lg x=0, \lg y=1$, then
$$
x=1, y=10, \lg \left(x+\frac{y}{x}\right)=\lg 11>1 \text {; }
$$

If $\lg x=1, \lg y=0$, then
$$
\lg \left(x+\frac{y}{x}\right)=\lg \left(10+\frac{1}{10}\right)<\lg 11 \text {; }
$$

If $\lg x=0, \lg \left(x+\frac{y}{x}\right)=1$, then $x=1, y=9, \lg y=\lg 9<\lg 11$;
If $\lg y=0, \lg \left(x+\frac{y}{x}\right)=1$, then $y=1, x+\frac{1}{x}=10, x<11, \lg x<\lg 11$;
If $\lg y=1, \lg \left(x+\frac{y}{x}\right)=0$, then $y=10, x+\frac{10}{x}=1$, which is impossible.
In summary, the maximum possible value of the largest element of $A$ is $\lg 11$.