Suppose that\begin{equation*} \lim_{x \to a^-} f(x) = L \ \ and \ \ \lim_{x \to a^-} g(x) = M\end{equation*}If $f(x) \leq g(x) \ \forall x$ in a deleted neighborhood of $a$, then $L \leq M$.
Proof. Assume for the sake of contradiction that $L > M$. Take $\epsilon = L - M$. Via the theorem of difference of limits, $\lim_{x \to a^-}[g(x) - f(x)] = M - L$. Therefore, for a given $\epsilon > 0$ there exists a $\delta > 0$ s.t.\begin{align*} -\delta < x-a < 0 &\implies |g(x) - f(x) - (M-L)| < \epsilon \\&\implies g(x) - f(x) - M + L < L - M \\&\implies g(x) - f(x) < 0 \\&\implies g(x) < f(x)\end{align*}
However, $g(x) \geq f(x)$ by assumption so we have reached a contradiction. Therefore, $L \leq M$.
Is this proof valid? I am trying to prove the analogs of the basic limit theorems. I simply change the typical $0 < |x-a| < \delta$ for either $-\delta < x-a < 0$ or $0 < x-a < \delta$ when dealing with left or right sided limits, respectively?