Question

Предел: lim x->0 (x)/(sqrt(3+x)-sqrt(3-x))​

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

To find the limit of $\frac{x}{\sqrt{3+x} - \sqrt{3-x}}$ as $x$ approaches 0, we can use a common technique called rationalizing the numerator. Here's how you can do it step by step:

1. Multiply both the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3+x} + \sqrt{3-x}$. This is done to eliminate the square root in the denominator:

$$\lim_{{x \to 0}} \frac{x}{\sqrt{3+x} - \sqrt{3-x}} \times \frac{\sqrt{3+x} + \sqrt{3-x}}{\sqrt{3+x} + \sqrt{3-x}}$$

2. Simplify the expression:

$$= \lim_{{x \to 0}} \frac{x (\sqrt{3+x} + \sqrt{3-x})}{(3+x) - (3-x)}$$

3. Simplify further:

$$= \lim_{{x \to 0}} \frac{x (\sqrt{3+x} + \sqrt{3-x})}{2x}$$

4. Cancel out $x$ in the numerator and denominator:

$$= \lim_{{x \to 0}} \frac{\sqrt{3+x} + \sqrt{3-x}}{2}$$

Now, as $x$ approaches 0, $\sqrt{3+x}$ approaches $\sqrt{3}$ and $\sqrt{3-x}$ also approaches $\sqrt{3}$. Therefore, the expression becomes:

$$= \frac{\sqrt{3} + \sqrt{3}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$$

So, the limit of $\frac{x}{\sqrt{3+x} - \sqrt{3-x}}$ as $x$ approaches 0 is $\sqrt{3}$.