Evaluate the following limits :
`Lim_( x to 0) (x^(2))/(1- cos x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{x^2}{1 - \cos x}, \]

we will follow these steps:

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the limit:

\[ \frac{0^2}{1 - \cos(0)} = \frac{0}{1 - 1} = \frac{0}{0}. \]

This results in an indeterminate form \( \frac{0}{0} \).

**Hint:** When you encounter an indeterminate form like \( \frac{0}{0} \), consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and the denominator separately.

Differentiate the numerator \( x^2 \):

\[ \frac{d}{dx}(x^2) = 2x. \]

Differentiate the denominator \( 1 - \cos x \):

\[ \frac{d}{dx}(1 - \cos x) = \sin x. \]

Now, we can rewrite the limit:

\[ \lim_{x \to 0} \frac{x^2}{1 - \cos x} = \lim_{x \to 0} \frac{2x}{\sin x}. \]

**Hint:** After applying L'Hôpital's Rule, check if the new limit still results in an indeterminate form.

### Step 3: Substitute \( x = 0 \) Again

Now we substitute \( x = 0 \) again:

\[ \frac{2(0)}{\sin(0)} = \frac{0}{0}. \]

We still have an indeterminate form \( \frac{0}{0} \).

**Hint:** If you still have an indeterminate form, you can apply L'Hôpital's Rule again.

### Step 4: Apply L'Hôpital's Rule Again

Differentiate the numerator \( 2x \):

\[ \frac{d}{dx}(2x) = 2. \]

Differentiate the denominator \( \sin x \):

\[ \frac{d}{dx}(\sin x) = \cos x. \]

Now, we can rewrite the limit again:

\[ \lim_{x \to 0} \frac{2x}{\sin x} = \lim_{x \to 0} \frac{2}{\cos x}. \]

**Hint:** After differentiating again, evaluate the limit by substituting \( x = 0 \).

### Step 5: Substitute \( x = 0 \) Once More

Now we substitute \( x = 0 \):

\[ \frac{2}{\cos(0)} = \frac{2}{1} = 2. \]

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{x^2}{1 - \cos x} = 2. \]

### Final Answer

The final answer is:

\[ \boxed{2}. \]