$$ \lim_{n-\infty}n\left(\sqrt{n\left(n-2\right)}-\sqrt{n^2-3}\right. $$

To evaluate the limit

\[
\lim_{n \to \infty} n \left( \sqrt{n(n-2)} - \sqrt{n^2 - 3} \right),
\]

we can simplify by rewriting the expression in a form that allows us to cancel out the terms in the square roots. Here’s how we can proceed:

### Step 1: Rewrite by Rationalizing
We have:
\[
n \left( \sqrt{n(n-2)} - \sqrt{n^2 - 3} \right).
\]
To simplify, we can multiply and divide by the conjugate of the square root expression:

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

This gives:
\[
= \frac{n \left( \left( \sqrt{n(n-2)} \right)^2 - \left( \sqrt{n^2 - 3} \right)^2 \right)}{\sqrt{n(n-2)} + \sqrt{n^2 - 3}}.
\]

### Step 2: Expand and Simplify the Numerator
Expanding the squares in the numerator, we get:
\[
= \frac{n \left( n(n-2) - (n^2 - 3) \right)}{\sqrt{n(n-2)} + \sqrt{n^2 - 3}}.
\]

Expanding $ n(n-2) $ and $ n^2 - 3 $ gives:
\[
= \frac{n \left( n^2 - 2n - n^2 + 3 \right)}{\sqrt{n(n-2)} + \sqrt{n^2 - 3}}.
\]

Simplifying further in the numerator:
\[
= \frac{n \left( -2n + 3 \right)}{\sqrt{n(n-2)} + \sqrt{n^2 - 3}}.
\]

Now the expression becomes:
\[
= \frac{-2n^2 + 3n}{\sqrt{n(n-2)} + \sqrt{n^2 - 3}}.
\]

### Step 3: Divide by $ n $ in the Denominator
As $ n \to \infty $, we can divide both the numerator and the denominator by $ n $ to simplify the limit:

\[
= \frac{-2n + 3}{\sqrt{n - 2} + \sqrt{n - \frac{3}{n}}}.
\]

### Step 4: Take the Limit as $ n \to \infty $
As $ n \to \infty $, the terms $ -2 $ and $ -\frac{3}{n} $ in the square roots become negligible, so we get:

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

Thus, the limit is:

\[
\lim_{n \to \infty} n \left( \sqrt{n(n-2)} - \sqrt{n^2 - 3} \right) = -\infty.
\]

Let me know if you need more details on any of the steps.

---

**Questions for Further Exploration:**
1. What happens if we consider similar limits but with different exponents in the square root?
2. How would this limit change if we replaced $n(n-2)$ with $n(n-1)$?
3. Can we use L'Hôpital's Rule to solve similar types of limits?
4. How would we approach the limit if both square roots contained terms with higher powers?
5. Are there alternative techniques to rationalization for simplifying such limits?

**Tip:** In limits with square roots, rationalizing by multiplying with the conjugate is often very helpful to simplify expressions.

lim_{n \to \infty} n \left( \sqrt{n(n-2)} - \sqrt{n^2 - 3} \right)

Learn how to evaluate the limit $ \lim_{n \to \infty} n ( \sqrt{n(n-2)} - \sqrt{n^2 - 3}) $ using rationalization and algebraic manipulation. This step-by-step solution covers limit theorems, square root simplifications, and conjugate multiplication techniques. Suitable for high school students in Grades 11-12.

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