We know that
\begin{align*} (2 + 3) (2^{-1} + 3^{-1}) &\equiv 1 \pmod{11} \\ 2 \cdot 2^{-1} + 2 \cdot 3^{-1} + 3 \cdot 2^{-1} + 3 \cdot 3^{-1} &\equiv 1 \pmod{11}, \end{align*}
or
\[5 \cdot (2^{-1} + 3^{-1}) \equiv 1 \pmod{11}.\]Therefore, the inverse of 5 modulo 11 is $2^{-1} + 3^{-1},$ which means $(a + b)^{-1} \equiv a^{-1} + b^{-1} \pmod{11}.$ Hence, $L - R = \boxed{0}.$
Given $m\geq 2$, denote by $b^{-1}$ the inverse of $b\pmod{m}$. That is, $b^{-1}$ is the residue for which $bb^{-1}\equiv 1\pmod{m}$. Sadie wonders if $(a+b)^{-1}$ is always congruent to $a^{-1}+b^{-1}$ (modulo $m$). She tries the example $a=2$, $b=3$, and $m=11$. Let $L$ be the residue of $(2+3)^{-1}\pmod{11}$, and let $R$ be the residue of $2^{-1}+3^{-1}\pmod{11}$, where $L$ and $R$ are integers from $0$ to $10$ (inclusive). Find $L-R$.
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