A well known fact is that $$\lim_{n\to\infty} \left(1+\frac{a}{n}\right)^n=e^a$$
Now I was wondering what if $a$ also depends on $n$? In particular take $$\lim_{n\to\infty} \left(1+\frac{a+cn}{n}\right)^n$$
is it possible to say that this behaves as $e^{a+cn}$ for large $n$?
With 'behaves' I mean something like: does $$\left(1+\frac{a+cn}{n}\right)^nb^{cn}$$ have the same limit (for $n\to\infty$) as $e^{a}(eb)^{cn}$?
Thanks for your help!
$(1+\frac{a+cn}{n})^n=(1+c+\frac{a}{n})^n=(1+c)^n(1+\frac{\frac{a}{1+c}}{n})^n\sim(1+c)^ne^\frac{a}{1+c}$ where $\sim$ means that the quotient of two sides tends to 1 as $n\rightarrow\infty$. Now if $c$ is close enough to 0, we have $1+c\approx e^c$, so we can write $(1+\frac{a+cn}{n})^n\approx e^{cn}\cdot e^\frac{a}{1+c}=e^{cn+\frac{a}{1+c}}$. Again, if $c\approx 0$, then $\frac{a}{1+c}\approx a$, so we can say that, in certain informal sense, $(1+\frac{a+cn}{n})^n\approx e^{cn+a}$.
As for existence of the limit, note that $\frac{(1+c)^ne^\frac{a}{1+c}}{e^{cn+a}}\frac{(1+c)^n}{(e^c)^n}\cdot\frac{e^\frac{a}{1+c}}{e^a}=(\frac{1+c}{e^c})^n\cdot e^{\frac{a}{1+c}-a}$. The limit of this expression can only exist and be non-zero if $\frac{1+c}{e^c}=1$, which is possible only for $c=0$. So as long as we have $c\neq 0$ then limit in question does not exist.