If the difference between two oracles is negligible, is the difference between a PPT algorithm with these two oracles also negligible?
Given $x$, define $p_1(x,y) = \Pr[O_1(x) = y]$ and $p_2(x,y) = \Pr[O_2(x) = y]$. Let
$$
\begin{align*}
\delta_x &amp;= \sum_y [\max(p_1(x,y),p_2(x,y)) - \min(p_1(x,y),p_2(x,y))] \\
&amp;= \sum_y |p_1(x,y) - p_2(x,y)| \\ &amp;\leq \epsilon_1(n).
\end{align*}
$$
On the other hand,
$$
\sum_y [\max(p_1(x,y),p_2(x,y)) + \min(p_1(x,y),p_2(x,y))] = \sum_y [p_1(x,y) + p_2(x,y)] = 2.
$$
Therefore
$$
\sum_y \min(p_1(x,y),p_2(x,y)) = 1 - \frac{\delta_x}{2}.
$$
Consider now the following random process that generates two random variables $R_1,R_2$:


For every $y$, with probability $\min(p_1(x,y),p_2(x,y))$, generate $R_1=R_2=y$.
Otherwise (this happens with probability $\delta_x/2$):


For every $y$, with probability $\frac{p_1(x,y)-\min(p_1(x,y),p_2(x,y))}{\delta_x/2}$ generate $R_1=y$.
For every $y$, with probability $\frac{p_2(x,y)-\min(p_1(x,y),p_2(x,y))}{\delta_x/2}$ generate $R_2=y$.



This process is a coupling that generates two random variables $R_1,R_2$ with the following properties:


$R_1 \sim O_1(x)$.
$R_2 \sim O_2(x)$.
$\Pr[R_1 \neq R_2] = \delta_x \leq \epsilon_1(n)$.




Let us go back to your algorithm $A$, and suppose that it runs in time $T(n)$. Use the coupling above to generate two random variables $B_1,B_2$ with the following properties:


$B_1 \sim A^{O_1}(x)$.
$B_2 \sim A^{O_2}(x)$.
$\Pr[B_1 \neq B_2] \leq T(n) \epsilon_1(n) =: \epsilon_2(n)$.


Note that $\epsilon_2(n)$ is negligible. We have

$$
\begin{align*}
&amp;\sum_y |\Pr[A^{O_1}(x) = y] - \Pr[A^{O_2}(x) = y]| \\ =
&amp;\sum_y |\Pr[B_1 = y] - \Pr[B_2 = y]| \\ =
&amp;\Pr[B_1 \neq B_2] \sum_y |\Pr[B_1=y|B_1 \neq B_2] - \Pr[B_2=y|B_1 \neq B_2]| \\ \leq 
&amp;\Pr[B_1 \neq B_2] \sum_y (\Pr[B_1=y|B_1 \neq B_2] + \Pr[B_2=y|B_1 \neq B_2]) \\ \leq &amp;2\epsilon_2(n),
\end{align*}
$$
since given $\Pr[B_1=y|B_1=B_2] = \Pr[B_2=y|B_1=B_2]$.