Evaluate the following limits :
`Lim_(x to 0 ) (tan (x+a) -tana)/(3x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\tan(x + a) - \tan(a)}{3x}, \]

we can follow these steps:

### Step 1: Use the formula for the difference of tangents
We know the identity for the difference of tangents:

\[ \tan(x + a) - \tan(a) = \frac{\sin(x)}{\cos(x + a) \cos(a)}. \]

However, for small values of \(x\), we can also use the Taylor series expansion for \(\tan(x)\) around 0, which gives us:

\[ \tan(x) \approx x + \frac{x^3}{3} + O(x^5). \]

### Step 2: Expand \(\tan(x + a)\)
Using the Taylor series, we can write:

\[ \tan(x + a) = \tan(a) + \sec^2(a) \cdot x + O(x^3). \]

Thus, we have:

\[ \tan(x + a) - \tan(a) = \sec^2(a) \cdot x + O(x^3). \]

### Step 3: Substitute back into the limit
Now substituting this back into our limit gives:

\[ \lim_{x \to 0} \frac{\sec^2(a) \cdot x + O(x^3)}{3x}. \]

### Step 4: Simplify the expression
We can simplify this limit:

\[ \lim_{x \to 0} \left(\frac{\sec^2(a) \cdot x}{3x} + \frac{O(x^3)}{3x}\right) = \lim_{x \to 0} \left(\frac{\sec^2(a)}{3} + \frac{O(x^2)}{3}\right). \]

As \(x\) approaches 0, the \(O(x^2)\) term will vanish, leading to:

\[ \frac{\sec^2(a)}{3}. \]

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

\[ \lim_{x \to 0} \frac{\tan(x + a) - \tan(a)}{3x} = \frac{\sec^2(a)}{3}. \]

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