3. Let $[x]$ denote the greatest integer not exceeding the real number $x$ (for example, $[3.1]=3,[-3.1]=-4$). Suppose the real number $x$ is not an integer, and $x+\frac{113}{x}=[x]+\frac{113}{[x]}$. Then the value of $x$ is
3. $-10 \frac{3}{11}$.

Eliminating the denominator, we get
$$
x^{2}[x]+113[x]=x[x]^{2}+113 x \text {. }
$$

Rearranging and factoring, we get
$$
(x-[x])(x[x]-113)=0 \text {. }
$$

Since $x$ is not an integer, $x-[x] \neq 0$.
Therefore, $x[x]-113=0$.
Let $x=[x]+t(0<t<1)$. Substituting into equation (1), we get
$$
[x]^{2}+t[x]-113=0 \text {. }
$$

Solving for $t$, we get $t=\frac{113-[x]^{2}}{[x]}$.
From $0<t<1$, it is easy to see that $[x]=-11, t=\frac{8}{11}$.
Thus, $x=-11+\frac{8}{11}=-10 \frac{3}{11}$.