`lim_(x to 0) x/(tan^(-1)(2x))` is equal to `1/2`

Answer

To solve the limit \( \lim_{x \to 0} \frac{x}{\tan^{-1}(2x)} \), we can use the Taylor series expansion for \( \tan^{-1}(x) \) around \( x = 0 \).

### Step-by-step Solution:

1. **Write the limit expression**:
\[ L = \lim_{x \to 0} \frac{x}{\tan^{-1}(2x)} \]

2. **Use the Taylor series expansion for \( \tan^{-1}(x) \)**:
The Taylor series expansion for \( \tan^{-1}(x) \) around \( x = 0 \) is:
\[ \tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \]
Substituting \( 2x \) into this expansion gives:
\[ \tan^{-1}(2x) = 2x - \frac{(2x)^3}{3} + \frac{(2x)^5}{5} - \cdots \]
Simplifying this, we have:
\[ \tan^{-1}(2x) = 2x - \frac{8x^3}{3} + \frac{32x^5}{5} - \cdots \]

3. **Substitute the expansion into the limit**:
Now substitute this back into the limit:
\[ L = \lim_{x \to 0} \frac{x}{2x - \frac{8x^3}{3} + \frac{32x^5}{5} - \cdots} \]

4. **Factor out \( x \) from the denominator**:
Factor \( x \) out of the denominator:
\[ L = \lim_{x \to 0} \frac{x}{x \left(2 - \frac{8x^2}{3} + \frac{32x^4}{5} - \cdots\right)} \]
This simplifies to:
\[ L = \lim_{x \to 0} \frac{1}{2 - \frac{8x^2}{3} + \frac{32x^4}{5} - \cdots} \]

5. **Evaluate the limit as \( x \to 0 \)**:
As \( x \to 0 \), the higher order terms vanish:
\[ L = \frac{1}{2 - 0} = \frac{1}{2} \]

6. **Conclusion**:
Therefore, we conclude that:
\[ \lim_{x \to 0} \frac{x}{\tan^{-1}(2x)} = \frac{1}{2} \]