https://file.aiquickdraw.com/mathbot/file/1729016071934xw4sdsbe.jpg

The expression in your image is the limit:

\[
\lim_{{n \to +\infty}} \frac{\sqrt{n+2} - 2}{\sqrt{n+7}} = 3
\]

Let's solve this step by step:

### Step 1: Simplify the expression
For large values of \(n\), the constants 2 and 7 become negligible compared to \(n\). So, both \(\sqrt{n+2}\) and \(\sqrt{n+7}\) behave approximately like \(\sqrt{n}\) for large \(n\). Let's explore this more formally by dividing both the numerator and the denominator by \(\sqrt{n}\):

\[
\frac{\sqrt{n+2} - 2}{\sqrt{n+7}} = \frac{\sqrt{n}\sqrt{1 + \frac{2}{n}} - 2}{\sqrt{n}\sqrt{1 + \frac{7}{n}}}
\]

### Step 2: Factor and simplify the expression for large \(n\)
As \(n \to \infty\), both \(\frac{2}{n}\) and \(\frac{7}{n}\) approach 0. Thus, we can use approximations for the square roots:
\[
\sqrt{1 + \frac{2}{n}} \approx 1 + \frac{1}{n}, \quad \sqrt{1 + \frac{7}{n}} \approx 1
\]

Substituting these approximations, the expression becomes:

\[
\frac{\sqrt{n}\left(1 + \frac{1}{n}\right) - 2}{\sqrt{n}} \approx \frac{\sqrt{n} + \frac{1}{\sqrt{n}} - 2}{\sqrt{n}}
\]

### Step 3: Separate terms
We can now separate the terms in the numerator:
\[
= \frac{\sqrt{n}}{\sqrt{n}} + \frac{\frac{1}{\sqrt{n}}}{\sqrt{n}} - \frac{2}{\sqrt{n}} = 1 + \frac{1}{n} - \frac{2}{\sqrt{n}}
\]

As \(n \to \infty\), the terms \(\frac{1}{n}\) and \(\frac{2}{\sqrt{n}}\) go to 0. Therefore, the expression approaches:

\[
1
\]

### Step 4: Conclusion
The limit is:

\[
\lim_{{n \to +\infty}} \frac{\sqrt{n+2} - 2}{\sqrt{n+7}} = 1
\]

It seems there was a mistake in the equation saying the limit equals 3. The actual limit is 1.

Do you want further clarification on any of the steps? 

Here are some related questions to expand on this topic:
1. How does the approximation of square roots work for large values of \(n\)?
2. What are some common techniques for finding limits involving square roots?
3. Can you apply L'Hopital's Rule to this type of limit?
4. How does dividing by the highest power of \(n\) help in limits?
5. What happens if you have different constants in the numerator and denominator?

**Tip**: When dealing with limits at infinity involving square roots, dividing by the largest term can greatly simplify the problem.

Limit of the expression (sqrt(n+2) - 2) / sqrt(n+7) as n approaches infinity.

Explore how to find the limit of (sqrt(n+2) - 2) / sqrt(n+7) as n approaches infinity. This detailed solution simplifies the expression using square root approximations and algebraic techniques. Suitable for undergraduate calculus students.

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