Limit of $e$ to the power of a difference of functions Suppose $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=\infty$ and $\lim_{x\to c}f(x)/g(x)=1$. Is it true that $$\lim_{x\to c}\frac{e^{f(x)}}{e^{g(x)}}=1$$ or not? What I've done so far: $$\lim_{x\to c}\frac{e^{f(x)}}{e^{g(x)}}=\lim_{x\to c}e^{f(x)-g(x)}=\exp\left\{\lim_{x\to c}[f(x)-g(x)]\right\}=\exp\left\{\lim_{x\to c}[f(x)-g(x)]\frac{g(x)}{g(x)}\right\}$$ Then, as $\lim_{x\to c}f(x)/g(x)=1$, $$\exp\left\{\lim_{x\to c}\left[\frac{f(x)}{g(x)}g(x)-g(x)\right]\right\}=\exp\left\{\lim_{x\to c}[g(x)-g(x)]\right\}=e^0=1$$ Is it correct or am I cheating when I introduce the $g(x)/g(x)$ term into the limit?
 A: You cannot solve this problem with the information you have. Because the exponential function is continuous the step
$$\lim_{x\to c}e^{f(x)-g(x)}=e^{\lim_{x\to c}(f(x)-g(x))}$$
is valid. But you don't know what the limit of the difference is. Your step 
$$
\lim_{x\to c}\left(\frac{f(x)}{g(x)}g(x)-g(x)\right)=\lim_{x\to c}(g(x)-g(x))
$$
is invalid, you've sort of half taken the limit here. The problem is that you don't know the behaviour of
$$
g(x)\left(1-\frac{f(x)}{g(x)}\right)
$$
as $x\to c$. One is going to infinity, the other to zero. Without knowledge of this you can't answer the question.
For example if $f(x)=g(x)=\frac1{x-c}$ then we get the right behaviour and the limit of the exponential of the difference is 1. But if we take $f(x)=\frac1{x-c}+1$ and $g(x)=\frac1{x-c}$ then we still have the right behaviour, but the limit of the difference of the exponential is $e$.
A: This preposition is not true. 
Let $f(x)=\frac1{|x-c|},\, g(x)=f(x)+\ln \pi$
$$\frac{e^{f(x)}}{e^{g(x)}}=e^{f(x)-g(x)}=e^{\ln \pi}=\pi,$$
and this is also your limit. Similar example can be constructed also for $c=\pm\infty.$
A: Suppose $f(x)=g(x)-h(x)$ then the condition is that the limit of $\frac {h(x)}{g(x)}\to 0$. If $g(x)\to \infty$ we could use $h(x)=\sqrt  {g(x)}$ to provide an example, where $h(x)$ also grows without limit.
Your limit then becomes the limit of $e^{h(x)}$, but as we have seen the growth of $h(x)$ is not sufficiently constrained to reach a conclusion here. Of course it is always possible to choose $h(x)=0$, so the limit $1$ is always possible.
