What is `lim_(xto0) (x)/(sqrt(1-cosx))` equal to ?

Options

A. `sqrt2`
B. `-sqrt2`
C. `(1)/(sqrt2)`
D. Limit does not exist

Explanation

To solve the limit \( \lim_{x \to 0} \frac{x}{\sqrt{1 - \cos x}} \), we can follow these steps:

### Step 1: Rewrite the cosine function
We start by using the trigonometric identity for cosine:
\[ \cos x = 1 - 2 \sin^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite \( 1 - \cos x \) as:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]

### Step 2: Substitute in the limit
Substituting this into our limit gives:
\[ \lim_{x \to 0} \frac{x}{\sqrt{1 - \cos x}} = \lim_{x \to 0} \frac{x}{\sqrt{2 \sin^2\left(\frac{x}{2}\right)}} \]
This simplifies to:
\[ \lim_{x \to 0} \frac{x}{\sqrt{2} \cdot \sin\left(\frac{x}{2}\right)} \]

### Step 3: Simplify the expression
Now we can factor out \( \sqrt{2} \):
\[ \lim_{x \to 0} \frac{1}{\sqrt{2}} \cdot \frac{x}{\sin\left(\frac{x}{2}\right)} \]

### Step 4: Use the limit property
We know from the standard limit property that:
\[ \lim_{u \to 0} \frac{u}{\sin u} = 1 \]
In our case, let \( u = \frac{x}{2} \), then as \( x \to 0 \), \( u \to 0 \) as well. Therefore, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{x}{\sin\left(\frac{x}{2}\right)} = \lim_{u \to 0} \frac{2u}{\sin u} = 2 \]

### Step 5: Combine the results
Now substituting back into our limit:
\[ \lim_{x \to 0} \frac{1}{\sqrt{2}} \cdot 2 = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to 0} \frac{x}{\sqrt{1 - \cos x}} = \sqrt{2} \]