Derivatives

Since the answer does not depend on $n$ we can make $n$ tend to $\infty$

So we have

$y= \sum_{0}^{\infty} \frac{x^{r}}{r!}$ as $n\to \infty$

So

$y=e^{x}$ as $n\to\infty$

So we have $\frac{dy}{dx} = e^{x}$ as $n\to \infty$

Also we have $\frac{x^{n}}{n!} \to 0$ as $n\to \infty$

To prove that we can use Stirling's Approximation

Or a simple theorem which states that

for a sequence $(x_{n})_{n}$ for which $\lim_{n\to \infty} \frac{x_{n+1}}{x_{n}} = k$ , $-1<k<1$ Then $(x_{n})_{n} \to 0$ as $n\to \infty$ .

So we have $\frac{dy}{dx} - y = e^{x} - e^{x} = 0$ as $n\to \infty$ .

This would be identically true as the answer does not depend on n .