Can I treat $\Delta t$ in different ways when $\lim_{\Delta t \to 0}$? i.e. $\Delta t = 0$ in one section and $\Delta t = d t$ in another? The Problem
Note: This is a possible solution to a question I asked a short time ago.
I am trying to prove that this sum
$$
\frac{1}{2 \sqrt{\pi}} \sum_{j=1}^{\infty} \sqrt{\frac{1}{j \Delta t}} (\cos{(\theta(t - j \Delta t - \Delta t))} - \cos{(\theta(t - j \Delta t + \Delta t))})$$
is exactly equivalent to this continuous function as $\lim_{\Delta t \to 0}$
$$
\sqrt{\frac{\theta}{2}} (\cos{(\theta t)} - \sin{(\theta t)})
$$
My Proof
I converted to complex notation and rearranged to get two sums, each of which is a polylogarithm function
$$
\frac{1}{4}\sqrt{\frac{1}{\pi \Delta t}}(e^{i \theta \Delta t} + e^{-i \theta \Delta t}) (e^{-i \theta \Delta t} \sum_{j=1}^{\infty} \frac{(e^{i \theta \Delta t})^j}{j^{\frac{1}{2}}} - e^{i \theta \Delta t} \sum_{j=1}^{\infty} \frac{(e^{-i \theta \Delta t})^j}{j^{\frac{1}{2}}})
$$
Both sums are solvable when $\lim_{\Delta t \to 0}$ (since the power term in the denominator is $<1$), giving the following
$$
\frac{1}{4}\sqrt{\frac{1}{\pi \Delta t}}(e^{i \theta \Delta t} + e^{-i \theta \Delta t})(e^{-i \theta t} \sqrt{\frac{i \pi}{\theta \Delta t}} - e^{i \theta t} \sqrt{\frac{\pi}{i \theta \Delta t}})
$$
And simplifying this gives
$$
\frac{1}{4 \Delta t}\sqrt{\frac{1}{\theta}}(e^{i \theta \Delta t} + e^{-i \theta \Delta t})(e^{-i \theta t} \sqrt{i} - e^{i \theta t} \sqrt{-i})
$$
The Question
To solve this, is it acceptable to simultaneously let $\Delta t = d t$ in the fraction and let $\Delta t = 0$ in the exponents? i.e. to yield
$$
\frac{1}{2 d t}\sqrt{\frac{1}{\theta}}(e^{-i \theta t} \sqrt{i} - e^{i \theta t} \sqrt{-i})
$$
so that I can then differentiate?
If so, using $e^{-i \theta t} \sqrt{i} + e^{i \theta t} \sqrt{-i} = \sqrt{2}(\sin(\theta t) + \cos{\theta t})$ to convert back to real numbers and then differentiating yields the continuous function above.
 A: As @Karl pointed out, it is not acceptable to let different instances of a variable "reach the limit at different times". On the suggestion of a colleague, I used the Taylor Expansion for $e^{i \theta \Delta t}+e^{-i \theta \Delta t}$ to remedy this error.
We can rewrite the sum as
$$
\frac{1}{4}\sqrt{\frac{1}{\pi}}\frac{(e^{i \theta \Delta t} - e^{-i \theta \Delta t})}{\Delta t}(e^{-i \theta t} \sqrt{\Delta t} \sum_{j=1}^{\infty} \frac{(e^{i \theta \Delta t})^j}{j^{\frac{1}{2}}} - e^{i \theta t} \sqrt{\Delta t} \sum_{j=1}^{\infty} \frac{(e^{-i \theta \Delta t})^j}{j^{\frac{1}{2}}})
$$
Now we use the following limits (where the summation terms are polylogarithms)
$$
\lim_{\Delta t \to 0} \frac{(e^{i \theta \Delta t} - e^{-i \theta \Delta t})}{\Delta t} = 2 i \theta
$$
$$
\lim_{\Delta t \to 0} \sqrt{\Delta t} \sum_{j=1}^{\infty} \frac{(e^{i \theta \Delta t})^j}{j^{\frac{1}{2}}} = \sqrt{\frac{i \pi}{\theta}}
$$
$$
\lim_{\Delta t \to 0} \sqrt{\Delta t} \sum_{j=1}^{\infty} \frac{(e^{-i \theta \Delta t})^j}{j^{\frac{1}{2}}} = \sqrt{\frac{\pi}{i \theta}}
$$
to derive
$$
\frac{\sqrt{\theta}}{2}i(e^{i \theta t} \sqrt{i} - e^{i \theta t} \sqrt{-1})
$$
Finally we convert back to trignometric form using $ i(e^{i \theta t} \sqrt{i} - e^{i \theta t} \sqrt{-1}) = \sqrt{2}(\sin{(\theta t)} - \cos{(\theta t))} $ to give
$$
\sqrt{\frac{\theta}{2}}(\sin{(\theta t)} - \cos{(\theta t}))
$$
However, note that this is NOT the same as the analytical solution; the sign is reversed. It appears that I missed a negative sign somewhere when deriving the original sum.
Thanks to @Karl and @Bannach for their critique of the original question.
