Math Problem Statement

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

Solution

To solve the limit

limn+n1+n3+n2+3n+1n3+2,\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 nn approaches infinity.

Step 1: Simplify the Square Roots

Let's examine the terms under the square roots. As nn \to \infty, the leading term in both square roots is n3n^3, which is dominant compared to the other terms. We can write:

n3+n2+3n+1n3=n3/2\sqrt{n^3 + n^2 + 3n + 1} \approx \sqrt{n^3} = n^{3/2} and n3+2n3=n3/2.\sqrt{n^3 + 2} \approx \sqrt{n^3} = n^{3/2}.

Thus, for large nn, both square roots tend to n3/2n^{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+n3+n2+3n+1+n3+2.1 + \sqrt{n^3 + n^2 + 3n + 1} + \sqrt{n^3 + 2}.

Multiplying by this conjugate gives:

limn+n(1+n3+n2+3n+1+n3+2)(1+n3+n2+3n+1n3+2)(1+n3+n2+3n+1+n3+2).\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+n3+n2+3n+1)2(n3+2)2.(1 + \sqrt{n^3 + n^2 + 3n + 1})^2 - (\sqrt{n^3 + 2})^2.

This difference of squares simplifies to:

1+(n3+n2+3n+1)(n3+2)=n2+3n1.1 + (n^3 + n^2 + 3n + 1) - (n^3 + 2) = n^2 + 3n - 1.

Step 3: Simplify the Expression

Now the limit becomes:

limn+n(1+n3+n2+3n+1+n3+2)n2+3n1.\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 nn, the dominant term in the denominator is n2n^2, so we can approximate:

limn+n(1+2n3/2)n2=limn+n2n3/2n2.\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:

=limn+2n2n2=2.= \lim_{{n \to +\infty}} \frac{2n^{2}}{n^2} = 2.

Conclusion

limn+n1+n3+n2+3n+1n3+2=2.\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?


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 nn \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 nn 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.

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Math Problem Analysis

Mathematical Concepts

Limits
Infinity
Rationalization
Asymptotic Behavior

Formulas

Difference of Squares
Square Root Approximation for Dominant Terms

Theorems

Limit Theorem for Infinity

Suitable Grade Level

Grades 11-12