If the value of `underset(x to 0)lim (sqrt(2+x)-sqrt2)/(x)" is equal to "1/(a sqrt2)` then 'a' equals

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the limit problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{\sqrt{2+x} - \sqrt{2}}{x} \]

### Step 1: Identify the form of the limit
When we substitute \(x = 0\) directly into the expression, we get:

\[ \frac{\sqrt{2+0} - \sqrt{2}}{0} = \frac{\sqrt{2} - \sqrt{2}}{0} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

**Hint:** Check the form of the limit by substituting the value of \(x\) before applying L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, if we have a limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can differentiate the numerator and the denominator separately.

Let:
- \(f(x) = \sqrt{2+x} - \sqrt{2}\)
- \(g(x) = x\)

Now, we differentiate \(f(x)\) and \(g(x)\):

1. The derivative of \(f(x)\):
\[ f'(x) = \frac{d}{dx}(\sqrt{2+x}) = \frac{1}{2\sqrt{2+x}} \cdot (1) = \frac{1}{2\sqrt{2+x}} \]

2. The derivative of \(g(x)\):
\[ g'(x) = \frac{d}{dx}(x) = 1 \]

Now, we apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{2+x}}}{1} = \lim_{x \to 0} \frac{1}{2\sqrt{2+x}} \]

**Hint:** Remember to differentiate both the numerator and denominator when using L'Hôpital's Rule.

### Step 3: Evaluate the limit
Now, substituting \(x = 0\):

\[ \lim_{x \to 0} \frac{1}{2\sqrt{2+x}} = \frac{1}{2\sqrt{2+0}} = \frac{1}{2\sqrt{2}} \]

### Step 4: Set the limit equal to the given expression
We are given that this limit equals \(\frac{1}{a\sqrt{2}}\):

\[ \frac{1}{2\sqrt{2}} = \frac{1}{a\sqrt{2}} \]

### Step 5: Solve for \(a\)
To find \(a\), we can cross-multiply:

\[ 1 \cdot a\sqrt{2} = 2\sqrt{2} \]

Dividing both sides by \(\sqrt{2}\):

\[ a = 2 \]

### Final Answer
Thus, the value of \(a\) is:

\[ \boxed{2} \]