$\displaystyle \lim_{x\to\infty}\left (a^x+b^x+c^x+d^x+...\right )^{\frac{1}{x}}$
$\displaystyle a^x$ will grow faster than all other terms. So we can ignore the others and conclude that,
$\displaystyle \lim_{x\to\infty}\left (a^x \right )^{\frac{1}{x}}$
$\displaystyle a$
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A more formal approach is here:
$\displaystyle \lim_{x\to\infty}\left (a^x+b^x+c^x+d^x+...\right )^{\frac{1}{x}}$
$\displaystyle \lim_{x\to\infty} e^{ \frac{1}{x}\ln(a^x+b^x+c^x+d^x+...)}$
Using L'hopital,
$\displaystyle \lim_{x\to\infty} e^{ \frac{a^x \ln a + b^x \ln b + c^x \ln c + d^x \ln d + ...}{a^x + b^x + c^x + d^x + ...} }$
Now divide all terms by $\displaystyle a^x$, then all terms except $\displaystyle a^x \ln a$ and $\displaystyle a^x$ will go to zero. So $\displaystyle e^{ \frac{a^x \ln a}{a^x} } = e^{\ln a} = a$
No takers? Oh well,
There are a couple of ways to do this.
The smart*** way would be that since
$\displaystyle a>b>c>d>...>0$
we have that $\displaystyle a^x+b^x+c^x+d^x+...\sim{a^x}$
This can be shown by the limit
$\displaystyle \lim_{x\to\infty}\frac{a^x+b^x+c^x+d^x+...}{a^x}=\ lim_{x\to\infty}\frac{1+\bigg(\frac{b}{a}\bigg)^x+ \bigg(\frac{c}{a}\bigg)^x+\bigg(\frac{d}{a}\bigg)^ x+...}{1}=1$
this was gotten from the fact that $\displaystyle \lim_{x\to\infty}\bigg(\frac{\gamma}{\delta}\bigg) ^x=0$ if $\displaystyle \delta>\gamma$
Therefore
$\displaystyle \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x+...\bigg)^{ \frac{1}{x}}\sim\lim_{x\to\infty}(a^x)^{\frac{1}{x }}=a$
The other answer that is better in my opinion would be the use of the ratio/root test trick
$\displaystyle \lim_{x\to\infty}\bigg(a^x+b^x+c^x+d^x\bigg)^{\fra c{1}{x}}=\lim_{x\to\infty}\frac{a\cdot{a^x}+b\cdot {b^x}+c\cdot{c^x}+d\cdot{d^x}+...}{a^x+b^x+c^x+d^x +...}$
Now by termwise division with $\displaystyle a^x$
we get
$\displaystyle \lim_{x\to\infty}\frac{a+b\bigg(\frac{b}{a}\bigg)^ x+c\bigg(\frac{c}{a}\bigg)^x+d\bigg(\frac{d}{a}\bi gg)^x+...}{1+\bigg(\frac{b}{a}\bigg)^x+\bigg(\frac {c}{a}\bigg)^x+\bigg(\frac{d}{a}\bigg)^x+...}=a$
EDITsorry, I am making these up as we go) Tis is another easy one, if someone would like to post another one I would be more than happy to give up the posting position
Find the smallest integer$\displaystyle \alpha$ such that
$\displaystyle \lim_{n\to\infty}\frac{(n!)^{\alpha}}{(3n)!}>1$
Ok, since no one is answering.
It cannot be one or two, since that would leave a factor of $\displaystyle n^n$ and $\displaystyle \alpha=3$ would still leave that pesky $\displaystyle 27^n$ in the denominator, but $\displaystyle \alpha=4$ does the trick
Does no one have any hard limits?
Not that you are incorrect in any way about this, but you are aware that this makes the limit go from 0 ($\displaystyle \alpha = 3$) to $\displaystyle \infty$ ($\displaystyle \alpha = 4$)?
I wonder if there is a way to find an $\displaystyle \alpha$ such that the limit is equal to 1?
-Dan