To solve the given limit

$\lim_{n \to \infty} \frac{(n+4)^2 - (n-2)^2}{\sqrt{(2n+1)^2 + (n+4)^2}},$

we'll simplify the numerator and denominator step by step.

Step 1: Expand the numerator

The numerator is $(n+4)^2 - (n-2)^2$. Expanding each term: $(n+4)^2 = n^2 + 8n + 16, \quad (n-2)^2 = n^2 - 4n + 4.$ Subtracting: $(n+4)^2 - (n-2)^2 = (n^2 + 8n + 16) - (n^2 - 4n + 4) = 12n + 12.$

So, the numerator simplifies to: $12n + 12 = 12(n + 1).$

Step 2: Simplify the denominator

The denominator is $\sqrt{(2n+1)^2 + (n+4)^2}$. Expanding each term: $(2n+1)^2 = 4n^2 + 4n + 1, \quad (n+4)^2 = n^2 + 8n + 16.$ Adding these: $(2n+1)^2 + (n+4)^2 = 4n^2 + 4n + 1 + n^2 + 8n + 16 = 5n^2 + 12n + 17.$

So, the denominator becomes: $\sqrt{5n^2 + 12n + 17}.$

Step 3: Express the limit

The limit becomes: $\lim_{n \to \infty} \frac{12(n+1)}{\sqrt{5n^2 + 12n + 17}}.$

Step 4: Factor $n^2$ from the square root

To handle the large $n$, factor $n^2$ out of the denominator: $\sqrt{5n^2 + 12n + 17} = \sqrt{n^2 \left(5 + \frac{12}{n} + \frac{17}{n^2}\right)} = n \sqrt{5 + \frac{12}{n} + \frac{17}{n^2}}.$

Step 5: Simplify the fraction

The fraction becomes: $\frac{12(n+1)}{\sqrt{5n^2 + 12n + 17}} = \frac{12(n+1)}{n \sqrt{5 + \frac{12}{n} + \frac{17}{n^2}}}.$

Divide numerator and denominator by $n$: $= \frac{12 \left(1 + \frac{1}{n}\right)}{\sqrt{5 + \frac{12}{n} + \frac{17}{n^2}}}.$

Step 6: Evaluate the limit as $n \to \infty$

As $n \to \infty$, the terms $\frac{1}{n}$, $\frac{12}{n}$, and $\frac{17}{n^2}$ all approach 0. Thus: $\sqrt{5 + \frac{12}{n} + \frac{17}{n^2}} \to \sqrt{5}.$ And: $1 + \frac{1}{n} \to 1.$

So, the limit becomes: $\frac{12 \cdot 1}{\sqrt{5}} = \frac{12}{\sqrt{5}}.$

Final Answer:

$\boxed{\frac{12}{\sqrt{5}}}.$

Let me know if you need further clarification or step-by-step verification!