For some analytic function $f(x)=\lim_{n\rightarrow\infty}\sum^n_{r=0}c_rx^r$, $$\lim_{x\rightarrow\infty}\frac{f(x)}{e^x}=\lim_{x\rightarrow\infty}\frac{\lim_{n\rightarrow\infty}\sum^n_{r=0}c_rx^r}{e^x}=\lim_{n\rightarrow\infty}\sum^n_{r=0}c_r\lim_{x\rightarrow\infty}\frac{x^r}{e^x}=\lim_{n\rightarrow\infty}\sum^n_{r=0}c_r(0)=\lim_{n\rightarrow\infty}0=0$$ But obviously this cannot be true? For say, $f(x)=e^x$ we would have $$\lim_{x\rightarrow\infty}\frac{e^x}{e^x}=1$$ and $$\lim_{x\rightarrow\infty}\frac{e^x}{e^x}=0$$ which would imply$$0=1$$
UPDATE: An answer has suggested that the two limits cannot be inverted. Is this true and why is this so? I would like a more complete explanation if this is the case. Also if this were the case, couldn't this modification to the question avoid that pitfall completely? $f(x)=\sum^\infty_{r=0}c_rx^r$, $$\lim_{x\rightarrow\infty}\frac{f(x)}{e^x}=\lim_{x\rightarrow\infty}\frac{\sum^\infty_{r=0}c_rx^r}{e^x}=\sum^\infty_{r=0}c_r\lim_{x\rightarrow\infty}\frac{x^r}{e^x}=\sum^\infty_{r=0}c_r(0)==0$$
Thanks in advance. Any other thoughts or comments welcome.
"Is this true?" Yes, and your example shows why it is true. The switching of limits leads to $0=1$, and therefore it is wrong.
"Why is this so?" Because the results of the limits in different orders is often different. You have given one example of this.
"Also if this were the case, couldn't this modification to the question avoid that pitfall completely?...." No, that actually is not a modification, just a more concise notation with the exact same meaning. The value of an infinite series is, by definition, the limit of the sequence of partial sums. That is, $\sum\limits_{r=0}^\infty a_r$ means the same thing as $\lim\limits_{n\to\infty}\sum\limits_{r=0}^n a_r$, provided the latter exists. Thus, passing a limit through an infinite summation is a case of switching limits, and as your example shows, it is not always correct.