How do you evaluate the limit of the numerator as x approaches infinity in the given expression?

Let's break down that step:

Step 1: Understanding the limit evaluation.
We had the expression for $N(x)$ as:

$$N(x) = \frac{2}{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}} + \sqrt{1+\frac{3}{x^2}}}$$

When we take the limit as $x \to \infty$, we consider what happens to each term involving $x$:

• For any constant $c$, $\lim_{x \to \infty} \frac{c}{x} = 0$.
• For any constant $c$, $\lim_{x \to \infty} \frac{c}{x^2} = 0$.

Applying this to our expression:

• $\lim_{x \to \infty} \frac{2}{x} = 0$
• $\lim_{x \to \infty} \frac{3}{x^2} = 0$ (in the first square root)
• $\lim_{x \to \infty} \frac{3}{x^2} = 0$ (in the second square root)

So, when we substitute these limit values into the expression for $N(x)$:

$$\lim_{x \to \infty} N(x) = \frac{2}{\sqrt{1+0+0} + \sqrt{1+0}}$$ $$= \frac{2}{\sqrt{1} + \sqrt{1}}$$ $$= \frac{2}{1 + 1}$$ $$= \frac{2}{2}$$ $$= 1$$

This is how we arrive at the value of 1 for the limit of the numerator.