@Aditya Garg. You can still go ahead with your proof that does not involve exponential or logarithmic functions. First, fix a $k$ to be determined later. Then, for all $n>2k$
\begin{aligned}
(1+x)^n\geq\binom{n}{k}x^k=\frac{n(n-1)\cdot\ldots\cdot(n-k+1)}{k!}x^k>\frac{n^k}{2^k}\frac{x^k}{k!}
\end{aligned}
This is because each factor in $n(n-1)\cdots(n-k+1)$ is of the form $n-j$ for $0\leq j\leq k<\frac{n}{2}$. From this, you obtained that
\begin{aligned}
\frac{n^x}{(1+x)^n}\leq \frac{2^kk!}{x^k}n^{x-k}
\end{aligned}
Then, you can see that by taking a fix $k$ with $k>x$ you get that
$$
\lim_{n\rightarrow\infty}\frac{n^x}{(1+x)^n}=0,\qquad (x>0)
$$
The argument is a beautiful one. You can replace $n^x$ by $n^a$ for a fix number $a$ and you get the notable limit
$$
\lim_{n\rightarrow\infty}\frac{n^a}{(1+x)^n}=0,\qquad (x>0)
$$
The advantage of this type of argument is that you do not need to define yet (properly) the exponential and/or logarithmic function and requires very few properties of numbers. Rudin's baby book page 57, has other notable limits that used this type of arguments.
Of course, if you have already gone through Calculus and just need to determine the limit, by all means, use the properties of the exponential, logarithms, L'Hospital rules, etc.
Here is another elementary answer that does not involve logarithms not exponentials. It only requires the fact that $a^n\xrightarrow{n\rightarrow\infty}0$ for any $0<a<1$.
Let $a_n=\frac{n^\alpha}{(1+p)^n}$, where $\alpha>0$ and $p>0$. Then
$$
\frac{a_{n+1}}{a_n}=\big(1+\frac{1}{n}\Big)^\alpha\frac{1}{1+p}
$$
Since $\Big(1+\frac{1}{n}\Big)^\alpha \xrightarrow{n\rightarrow\infty}1$, there is $N\in\mathbb{N}$ such that $n\geq N$ implies that $\big(1+\frac{1}{n}\Big)^\alpha<1+\frac{p}{2}$. Let
Then, for all $n\geq N$
$$a_{n+1}\leq \frac{2+p}{2+2p} a_n$$
This implies that
$$
a_{n+N}\leq \Big(\frac{2+p}{2+2p}\Big)^na_N\xrightarrow{n\rightarrow\infty}0
$$
since $0<\frac{2+p}{2+2p}<1$.
