$$ \lim _ { x \rightarrow + \infty } \sqrt { x + 4 - \sqrt { x } } $$

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1. Rewrite the expression:
   
   $$
   \sqrt{x+4-\sqrt{x}} = \sqrt{x\left(1 - \frac{1}{\sqrt{x}} + \frac{4}{x}\right)}.
   $$
   
2. Factor out $$\sqrt{x}$$ from the radical:
   
   $$
   \sqrt{x+4-\sqrt{x}} = \sqrt{x} \, \sqrt{1 - \frac{1}{\sqrt{x}} + \frac{4}{x}}.
   $$
   
3. Evaluate the limit of the second square root as $$x \to +\infty$$:
   
   As $$x \to +\infty$$,
   
   $$
   \frac{1}{\sqrt{x}} \to 0 \quad \text{and} \quad \frac{4}{x} \to 0.
   $$
   
   Therefore,
   
   $$
   \sqrt{1 - \frac{1}{\sqrt{x}} + \frac{4}{x}} \to \sqrt{1} = 1.
   $$
   
4. Now, the limit becomes:
   
   $$
   \lim_{x \to +\infty} \sqrt{x+4-\sqrt{x}} = \lim_{x \to +\infty} \sqrt{x} \cdot 1 = \lim_{x \to +\infty} \sqrt{x}.
   $$
   
5. Since $$\sqrt{x}$$ tends to $$+\infty$$ as $$x$$ tends to $$+\infty$$, we conclude:
   
   $$
   \lim_{x \to +\infty} \sqrt{x+4-\sqrt{x}} = +\infty.
   $$
The limit of $$ \sqrt{x + 4 - \sqrt{x}} $$ as $$ x $$ approaches $$ +\infty $$ is $$ +\infty $$.

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