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

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} \), we can follow these steps:

### Step 1: Use the identity for \( 1 - \cos x \)
We know the trigonometric identity:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x^2} \]

### Step 2: Rewrite the denominator
Next, we can rewrite \( x^2 \) in terms of \( \left(\frac{x}{2}\right)^2 \):
\[ x^2 = 4\left(\frac{x}{2}\right)^2 \]
Substituting this into our limit gives:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{4\left(\frac{x}{2}\right)^2} \]

### Step 3: Simplify the expression
This simplifies to:
\[ \lim_{x \to 0} \frac{2}{4} \cdot \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} = \lim_{x \to 0} \frac{1}{2} \cdot \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} \]

### Step 4: Apply the limit property
Using the limit property \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \), where \( u = \frac{x}{2} \):
\[ \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} = 1 \]
Thus,
\[ \lim_{x \to 0} \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} = 1 \]

### Step 5: Final calculation
Putting it all together, we have:
\[ \lim_{x \to 0} \frac{1}{2} \cdot 1 = \frac{1}{2} \]

### Final Answer
Thus, the limit evaluates to:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]

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