Math Problem Statement

Use the epsilon definition to find and prove lim n to infinity ((n^2+(square root n))/(n^2 +1))

Solution

To prove

limnn2+nn2+1=1\lim_{{n \to \infty}} \frac{n^2 + \sqrt{n}}{n^2 + 1} = 1

using the epsilon (ε\varepsilon) definition of limits, we need to show that for any ε>0\varepsilon > 0, there exists a positive integer NN such that for all n>Nn > N,

n2+nn2+11<ε.\left| \frac{n^2 + \sqrt{n}}{n^2 + 1} - 1 \right| < \varepsilon.

Step 1: Simplify the Expression

Let's start by simplifying the expression inside the absolute value:

n2+nn2+11=n2+n(n2+1)n2+1=n2+nn21n2+1=n1n2+1.\frac{n^2 + \sqrt{n}}{n^2 + 1} - 1 = \frac{n^2 + \sqrt{n} - (n^2 + 1)}{n^2 + 1} = \frac{n^2 + \sqrt{n} - n^2 - 1}{n^2 + 1} = \frac{\sqrt{n} - 1}{n^2 + 1}.

So, we need to make the expression n1n2+1\left|\frac{\sqrt{n} - 1}{n^2 + 1}\right| smaller than ε\varepsilon.

Step 2: Analyze the Simplified Expression as nn \to \infty

As nn becomes very large:

  • The denominator n2+1n^2 + 1 grows approximately like n2n^2.
  • The numerator n1\sqrt{n} - 1 grows much more slowly, since n=n1/2\sqrt{n} = n^{1/2}.

Thus, as nn \to \infty, n1n2+10\frac{\sqrt{n} - 1}{n^2 + 1} \to 0, suggesting that we can find an NN large enough for the expression to be less than any positive ε\varepsilon.

Step 3: Make the Inequality Precise

To satisfy n1n2+1<ε\left|\frac{\sqrt{n} - 1}{n^2 + 1}\right| < \varepsilon, we can bound the expression as follows:

n1n2+1nn2.\left|\frac{\sqrt{n} - 1}{n^2 + 1}\right| \leq \frac{\sqrt{n}}{n^2}.

Now simplify this expression further:

nn2=1n3/2.\frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}.

Step 4: Choose NN to Satisfy the Epsilon Condition

We want 1n3/2<ε\frac{1}{n^{3/2}} < \varepsilon. This inequality is satisfied if nn is large enough, specifically if:

n>(1ε)2/3.n > \left(\frac{1}{\varepsilon}\right)^{2/3}.

Let N=(1ε)2/3N = \left(\frac{1}{\varepsilon}\right)^{2/3}. Then for any n>Nn > N, we have:

n2+nn2+11<ε.\left|\frac{n^2 + \sqrt{n}}{n^2 + 1} - 1\right| < \varepsilon.

Conclusion

By the epsilon definition of a limit, we have shown that for any ε>0\varepsilon > 0, there exists an N=(1ε)2/3N = \left(\frac{1}{\varepsilon}\right)^{2/3} such that for all n>Nn > N,

n2+nn2+11<ε.\left| \frac{n^2 + \sqrt{n}}{n^2 + 1} - 1 \right| < \varepsilon.

Thus,

limnn2+nn2+1=1.\lim_{{n \to \infty}} \frac{n^2 + \sqrt{n}}{n^2 + 1} = 1.

Would you like more details on any specific part of this proof?


Here are five questions for further exploration:

  1. How can we apply the epsilon-delta definition for limits involving functions of xx rather than sequences?
  2. What would change if the numerator had a higher power term, like n3n^3 instead of n2n^2?
  3. How does the epsilon proof change for a limit approaching a non-infinity value?
  4. What are other ways to analyze limits involving square roots?
  5. How does this approach apply to limits in calculus of continuous functions?

Tip: Always examine both numerator and denominator growth rates to simplify limit calculations involving sequences.

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Math Problem Analysis

Mathematical Concepts

Limit of Sequences
Epsilon-Delta Definition of Limits
Asymptotic Behavior
Square Roots

Formulas

lim_{n->∞} f(n) = L
|f(n) - L| < ε

Theorems

Epsilon-Delta Definition of Limits

Suitable Grade Level

Grades 11-12