limit n tends to infinity for arbitrary non negative real numbers 
I tried evaluating the limit taking log and afterwards i couldnt proceed further. I tried solving using numbers instead of variables and did not got any relation from that and help in this matter is appreciated

Here is a version with L'Hôpital's rule.  It's not quite as strong (logically) as the other answers, as we'll explain.
Assume that $a_1 \leq a_2 \leq \dots \leq a_k$.  Starting from
$$
    L = \lim_{n\to\infty} \left(a_1^n + a_2^n + \dots + a_k^n\right)^{1/n}
$$
we know
$$\begin{split}
    \ln L &= \ln \lim_{n\to\infty} \left(a_1^n + a_2^n + \dots + a_k^n\right)^{1/n} \\
    &= \lim_{n\to\infty}\ln\left(a_1^n + a_2^n + \dots + a_k^n\right)^{1/n} \\
    &= \lim_{n\to\infty}\frac{\ln\left(a_1^n + a_2^n + \dots + a_k^n\right)}{n}\\
    &\stackrel{\text{H}}{=} \lim_{n\to\infty}\frac{(\ln a_1) a_1^n + (\ln a_2)a_2^n + \dots + (\ln a_k)a_k^n}{a_1^n + a_2^n + \dots + a_k^n} \\
    &= \lim_{n\to\infty} \frac{a_k^n}{a_k^n}\cdot \frac{(\ln a_1) \left(\frac{a_1}{a_k}\right)^n + (\ln a_2)\left(\frac{a_2}{a_k}\right)^n + \dots + (\ln a_{k-1})\left(\frac{a_{k-1}}{a_k}\right)^n+ \ln a_k}{\left(\frac{a_1}{a_k}\right)^n + \left(\frac{a_2}{a_k}\right)^n + \dots + \left(\frac{a_{k-1}}{a_k}\right)^n+1} \\
&= 1 \cdot \frac{0 + 0 + \dots + 0 + \ln a_k}{0 + 0 + \dots + 0 + 1} = \ln a_k
\end{split}$$
(“H” denotes the invocation of L'Hôpital's rule).  Therefore $L=a_k$.
Remarks:


*

*Limit proofs with L'Hôpital's rule are “backwards” in the sense that we don't know that the limit $L$ exists until we get to the end.

*To verify the power sequence is indeterminate requires a check of cases.  If $a_k > 1$ then the base $a_1^n + \dots + a_k^n$ tends to $\infty$ so the limit is of the indeterminate form $\infty^0$.  If $a_k < 1$ then the base $a_1^n + \dots + a_k^n$ tends to $0$ so the limit is of the indeterminate form $0^0$.  If $a_k =1$, the base $a_1^n + \dots + a_k^n$ tends to $1$.  This form ($1^0$) is not indeterminate; the limit is $1$, but that requires a separate proof without L'Hôpital's rule.
