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To solve the limit

\[
\lim_{{n \to +\infty}} \frac{\sqrt{n}}{1 + \sqrt{n^3 + n^2 + 3n + 1} - \sqrt{n^3 + 2}},
\]

we'll go through a few steps to simplify the expression, especially as \( n \) approaches infinity.

### Step 1: Simplify the Square Roots
Let's examine the terms under the square roots. As \( n \to \infty \), the leading term in both square roots is \( n^3 \), which is dominant compared to the other terms. We can write:

\[
\sqrt{n^3 + n^2 + 3n + 1} \approx \sqrt{n^3} = n^{3/2}
\]
and
\[
\sqrt{n^3 + 2} \approx \sqrt{n^3} = n^{3/2}.
\]

Thus, for large \( n \), both square roots tend to \( n^{3/2} \), and the difference between them becomes small. 

### Step 2: Rationalize the Denominator
To handle the expression more effectively, we can rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator. The conjugate is:

\[
1 + \sqrt{n^3 + n^2 + 3n + 1} + \sqrt{n^3 + 2}.
\]

Multiplying by this conjugate gives:

\[
\lim_{{n \to +\infty}} \frac{\sqrt{n} \cdot (1 + \sqrt{n^3 + n^2 + 3n + 1} + \sqrt{n^3 + 2})}{\left(1 + \sqrt{n^3 + n^2 + 3n + 1} - \sqrt{n^3 + 2}\right) \cdot \left(1 + \sqrt{n^3 + n^2 + 3n + 1} + \sqrt{n^3 + 2}\right)}.
\]

Expanding the denominator:

\[
(1 + \sqrt{n^3 + n^2 + 3n + 1})^2 - (\sqrt{n^3 + 2})^2.
\]

This difference of squares simplifies to:

\[
1 + (n^3 + n^2 + 3n + 1) - (n^3 + 2) = n^2 + 3n - 1.
\]

### Step 3: Simplify the Expression
Now the limit becomes:

\[
\lim_{{n \to +\infty}} \frac{\sqrt{n} \cdot (1 + \sqrt{n^3 + n^2 + 3n + 1} + \sqrt{n^3 + 2})}{n^2 + 3n - 1}.
\]

For large \( n \), the dominant term in the denominator is \( n^2 \), so we can approximate:

\[
\lim_{{n \to +\infty}} \frac{\sqrt{n} \cdot (1 + 2n^{3/2})}{n^2} = \lim_{{n \to +\infty}} \frac{\sqrt{n} \cdot 2n^{3/2}}{n^2}.
\]

Simplifying the terms:

\[
= \lim_{{n \to +\infty}} \frac{2n^{2}}{n^2} = 2.
\]

### Conclusion
\[
\lim_{{n \to +\infty}} \frac{\sqrt{n}}{1 + \sqrt{n^3 + n^2 + 3n + 1} - \sqrt{n^3 + 2}} = 2.
\]

Would you like further clarification on any part of this solution?

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Here are some related questions to deepen your understanding:
1. How would the solution change if the dominant terms were different?
2. What happens if we do not use rationalization? Is there another method?
3. How can we determine when to approximate expressions as \( n \to \infty \)?
4. Why is the difference of squares technique useful for limits with square roots?
5. What would the result be if the exponent of \( n \) inside the square roots were different?

**Tip:** When dealing with limits involving square roots at infinity, rationalizing the denominator is often a very effective strategy.

Evaluate \( \lim_{{n \to +\infty}} \frac{\sqrt{n}}{1 + \sqrt{n^3 + n^2 + 3n + 1} - \sqrt{n^3 + 2}} \).

This solution provides a step-by-step evaluation of the limit \( \lim_{{n \to +\infty}} \frac{\sqrt{n}}{1 + \sqrt{n^3 + n^2 + 3n + 1} - \sqrt{n^3 + 2}} \) using techniques such as rationalization and approximation for dominant terms, suitable for advanced high school students.

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