How to demonstrate that $\frac{\sqrt{n}}{1+\sqrt{nx}}$ converges in $L^{1}( (0,1),\mu)?$

Let, $\mu$ be the Leb. measure. Let, $g_n =\frac{\sqrt{n}}{1+\sqrt{nx}} $ for $n \ge 1$. I know that,

$$\int^{1}_{0}|g_n| ~ d\mu \le \int^{1}_{0}\frac{\sqrt{n}}{\sqrt{nx}} ~ d\mu \le \int^{1}_{0}\frac{1}{\sqrt{x}} ~d\mu < \infty.$$ Thus, we can use the Lesbegue Dominated Convergence Theorem in this context. I know that,

$$\lim_{n\rightarrow \infty}g_n = \frac{1}{\sqrt{x}}.$$

I don't know where to go from here, any help would be appreciated. Thanks!

You have done most of the work. You want to show that $$\lim_{n \rightarrow \infty}\int_0^1 \bigg|\frac{\sqrt{n}}{1+\sqrt{nx}} - {1 \over \sqrt{x}}\bigg| = 0$$ You need a dominating function to apply the dominated convergence theorem here. Once you do that, you can put the limit on the inside, and based on what you have shown you will get the limit is zero and you have the desired $L^1$ convergence.

It will not be too hard to use the dominating function you've already constructed to get an appropriate dominating function for this limit.

Here is another way to do this problem.

You proved that $||g_{n}||_{1}$ are uniformly bounded, then it follows that $\{g_{n}\}$ is uniformly absolutely continuous. $g_{n}$ converges to $f(x)=\frac{1}{\sqrt{x}}$ in measure since $g_{n}$ converges pointwise to $f$ and the space is of finite measure. By Vitali's convergence theorem, $g_{n}$ also converges in $L^{1}$.