Evaluate the following limits:
`lim_(xrarr0)((tan2x-sin2x))/(x^(3))`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{\tan(2x) - \sin(2x)}{x^3} \), we will follow these steps:

### Step 1: Rewrite the limit
We start with the expression:
\[ \lim_{x \to 0} \frac{\tan(2x) - \sin(2x)}{x^3} \]

### Step 2: Use the Taylor series expansions
We can use the Taylor series expansions for \(\tan(2x)\) and \(\sin(2x)\) around \(x = 0\):
- The Taylor series for \(\tan(2x)\) is:
\[ \tan(2x) = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5) \]
- The Taylor series for \(\sin(2x)\) is:
\[ \sin(2x) = 2x - \frac{(2x)^3}{6} + O(x^5) = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5) \]

### Step 3: Substitute the expansions into the limit
Now substituting these expansions into the limit:
\[ \tan(2x) - \sin(2x) = \left(2x + \frac{8x^3}{3} + O(x^5)\right) - \left(2x - \frac{4x^3}{3} + O(x^5)\right) \]
This simplifies to:
\[ \tan(2x) - \sin(2x) = \frac{8x^3}{3} + \frac{4x^3}{3} + O(x^5) = \frac{12x^3}{3} + O(x^5) = 4x^3 + O(x^5) \]

### Step 4: Substitute back into the limit
Now we substitute this back into the limit:
\[ \lim_{x \to 0} \frac{4x^3 + O(x^5)}{x^3} \]
This gives:
\[ \lim_{x \to 0} \left(4 + \frac{O(x^5)}{x^3}\right) \]
As \(x\) approaches 0, \(\frac{O(x^5)}{x^3}\) approaches 0.

### Step 5: Evaluate the limit
Thus, we have:
\[ \lim_{x \to 0} \left(4 + 0\right) = 4 \]

### Final Answer
The limit is:
\[ \boxed{4} \]