Evaluate the following :
`lim_(x to 0)(1-cos^(2)x)/(3 tan^(2)x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{1 - \cos^2 x}{3 \tan^2 x}, \]

we can follow these steps:

### Step 1: Simplify the expression using trigonometric identities
We know that

\[ 1 - \cos^2 x = \sin^2 x. \]

So, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\sin^2 x}{3 \tan^2 x}. \]

### Step 2: Rewrite \(\tan^2 x\)
Recall that

\[ \tan x = \frac{\sin x}{\cos x}, \]

thus,

\[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x}. \]

Substituting this into our limit gives:

\[ \lim_{x \to 0} \frac{\sin^2 x}{3 \cdot \frac{\sin^2 x}{\cos^2 x}} = \lim_{x \to 0} \frac{\sin^2 x \cdot \cos^2 x}{3 \sin^2 x}. \]

### Step 3: Cancel out \(\sin^2 x\)
Assuming \(x \neq 0\), we can cancel \(\sin^2 x\) from the numerator and denominator:

\[ \lim_{x \to 0} \frac{\cos^2 x}{3}. \]

### Step 4: Evaluate the limit
As \(x\) approaches 0, \(\cos^2 x\) approaches \(\cos^2(0) = 1\). Therefore, we have:

\[ \lim_{x \to 0} \frac{\cos^2 x}{3} = \frac{1}{3}. \]

### Final Answer
Thus, the limit evaluates to:

\[ \frac{1}{3}. \]

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