Evaluate the following limits:
`lim_(xrarr0)((1-cosx))/(sin^(2)x)`

Explanation

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

### Step 1: Rewrite \(\sin^2 x\)
We know that \(\sin^2 x\) can be rewritten using the identity:
\[ \sin^2 x = 1 - \cos^2 x \]
Thus, we can express the limit as:
\[ \lim_{x \to 0} \frac{1 - \cos x}{1 - \cos^2 x} \]

### Step 2: Factor the denominator
Using the difference of squares, we can factor \(1 - \cos^2 x\):
\[ 1 - \cos^2 x = (1 - \cos x)(1 + \cos x) \]
Now, substituting this back into the limit gives us:
\[ \lim_{x \to 0} \frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)} \]

### Step 3: Cancel the common terms
Since \(1 - \cos x\) is common in the numerator and denominator (and is not zero as \(x\) approaches 0), we can cancel it out:
\[ \lim_{x \to 0} \frac{1}{1 + \cos x} \]

### Step 4: Evaluate the limit
Now we can directly substitute \(x = 0\) into the limit:
\[ 1 + \cos(0) = 1 + 1 = 2 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{1}{1 + \cos x} = \frac{1}{2} \]

### Final Answer
Therefore, the limit is:
\[ \frac{1}{2} \]

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