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

Answer

To evaluate the limit

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

we will use Taylor series expansions for \(e^x\) and \(\cos(2x)\) around \(x = 0\).

### Step 1: Expand \(e^x\) and \(\cos(2x)\)

The Taylor series expansion for \(e^x\) around \(x = 0\) is:

\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4). \]

Thus,

\[ e^x - 1 = x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4). \]

The Taylor series expansion for \(\cos(2x)\) around \(x = 0\) is:

\[ \cos(2x) = 1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} + O(x^6) = 1 - 2x^2 + \frac{4x^4}{24} + O(x^6). \]

Thus,

\[ 1 - \cos(2x) = 2x^2 - \frac{4x^4}{24} + O(x^6) = 2x^2 - \frac{x^4}{6} + O(x^6). \]

### Step 2: Substitute the expansions into the limit

Now we substitute these expansions into our limit:

\[ \lim_{x \to 0} \frac{x\left(x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right)}{2x^2 - \frac{x^4}{6} + O(x^6)}. \]

This simplifies to:

\[ \lim_{x \to 0} \frac{x^2 + \frac{x^3}{2} + \frac{x^4}{6} + O(x^5)}{2x^2 - \frac{x^4}{6} + O(x^6)}. \]

### Step 3: Factor out \(x^2\) from the numerator and denominator

Factoring \(x^2\) from both the numerator and denominator gives us:

\[ \lim_{x \to 0} \frac{x^2\left(1 + \frac{x}{2} + \frac{x^2}{6} + O(x^3)\right)}{x^2\left(2 - \frac{x^2}{6} + O(x^4)\right)}. \]

Cancelling \(x^2\) (valid since \(x \to 0\) and \(x^2 \neq 0\) for \(x \neq 0\)):

\[ \lim_{x \to 0} \frac{1 + \frac{x}{2} + \frac{x^2}{6} + O(x^3)}{2 - \frac{x^2}{6} + O(x^4)}. \]

### Step 4: Evaluate the limit as \(x \to 0\)

As \(x\) approaches 0, the terms involving \(x\) vanish:

\[ \frac{1 + 0 + 0}{2 - 0} = \frac{1}{2}. \]

Thus, the limit evaluates to:

\[ \boxed{\frac{1}{2}}. \]