$X$ is a random variable with PDF $f$, with $f$ continuous at $x=a$. Then $lim_{varepsilonto0^-}frac{1}{varepsilon}P(Xin(a,a+varepsilon))=f(a)$

The Problem: Let $X$ be a random variable with density function $f$, and assume that $f$ is continuous at $x=a$. Prove that
$$lim_{varepsilonsearrow0}frac{1}{varepsilon}P(Xin(a,a+varepsilon))=f(a).$$
My Attempt: Let $eta>0$ be given. Since $f$ is continuous at $a$, it also right-continuous at $a$. Therefore, we can choose a $delta>0$ such that if $0<varepsilon<delta$ then $vert f(a+varepsilon)-f(a)vert<eta.$ This also implies that if $0<x-a<delta$ then $vert f(x)-f(a)vert<eta.$ It follows that if $0<varepsilon<eta$, then
begin{align}
leftvertfrac{1}{varepsilon}P(Xin(a,a+varepsilon))-f(a)rightvert&=leftvertfrac{1}{varepsilon}int_a^{a+varepsilon}[f(x)-f(a),dxrightvert\
&leqfrac{1}{varepsilon}int_a^{a+varepsilon}vert f(x)-f(a)vert,dx\
&<frac{1}{varepsilon}int_a^{a+varepsilon}eta,dx\
&=eta.
end{align}
Therefore,
$$lim_{varepsilonsearrow0}frac{1}{varepsilon}P(Xin(a,a+varepsilon))=f(a).$$

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Do you agree with my approach and execution above?
Thank you very much for your time and appreciate any and all feedback.