https://file.aiquickdraw.com/mathbot/file/1725632432412xxsxhmwz.jpg

The expression given in the image is:

\[
\lim_{{x \to 0}} \frac{2e^x - \sqrt{\cos x}}{\sin x + \sin 2x}
\]

To solve this limit, we will use the Taylor series expansion of the functions around \(x = 0\). 

### Step-by-Step Solution:

1. **Expand Taylor Series Around \(x = 0\):**

   - For \(e^x\):
     \[
     e^x = 1 + x + \frac{x^2}{2} + O(x^3)
     \]

   - For \(\cos x\):
     \[
     \cos x = 1 - \frac{x^2}{2} + O(x^4)
     \]

   - For \(\sqrt{\cos x}\):
     \[
     \sqrt{\cos x} = \sqrt{1 - \frac{x^2}{2}} \approx 1 - \frac{x^2}{4} + O(x^4)
     \]

   - For \(\sin x\):
     \[
     \sin x = x - \frac{x^3}{6} + O(x^5)
     \]

   - For \(\sin 2x = 2 \sin x \cos x\):
     \[
     \sin 2x = 2(x - \frac{x^3}{6} + O(x^5))(1 - \frac{x^2}{2} + O(x^4)) \approx 2x - \frac{2x^3}{3} + O(x^5)
     \]

2. **Substitute Taylor Expansions in the Limit Expression:**

   Substitute the Taylor series expansions into the limit expression:

   \[
   2e^x - \sqrt{\cos x} \approx 2 \left(1 + x + \frac{x^2}{2}\right) - \left(1 - \frac{x^2}{4}\right)
   \]

   Simplify the numerator:

   \[
   2e^x - \sqrt{\cos x} \approx 2 + 2x + x^2 - 1 + \frac{x^2}{4} = 1 + 2x + \frac{5x^2}{4} + O(x^3)
   \]

   Simplify the denominator:

   \[
   \sin x + \sin 2x \approx \left(x - \frac{x^3}{6}\right) + \left(2x - \frac{2x^3}{3}\right)
   \]

   \[
   = 3x - \frac{x^3}{2} + O(x^5)
   \]

3. **Calculate the Limit:**

   Substitute these into the limit:

   \[
   \lim_{{x \to 0}} \frac{1 + 2x + \frac{5x^2}{4}}{3x - \frac{x^3}{2}}
   \]

   As \(x \to 0\), the dominant term in the denominator is \(3x\), so:

   \[
   \lim_{{x \to 0}} \frac{1 + 2x + \frac{5x^2}{4}}{3x} = \lim_{{x \to 0}} \left( \frac{1}{3x} + \frac{2}{3} + \frac{5x}{12} \right)
   \]

   The term \(\frac{1}{3x}\) goes to infinity as \(x \to 0\), hence the limit does not exist.

### Conclusion:

The limit diverges to infinity as \(x \to 0\). Thus, the result is:

\[
\lim_{{x \to 0}} \frac{2e^x - \sqrt{\cos x}}{\sin x + \sin 2x} = \infty
\]

Would you like more details, or do you have any questions?

### Follow-Up Questions:
1. What is the importance of Taylor series in solving limits?
2. How do we determine which terms to keep in Taylor expansions?
3. What happens if we use L'Hopital's rule for this limit?
4. How can we evaluate limits that tend to infinity?
5. What are some common mistakes when applying Taylor expansions in limits?

**Tip:** When evaluating limits involving indeterminate forms, consider both Taylor series expansion and L'Hopital's rule to find the simplest approach.

lim (x → 0) (2e^x - sqrt(cos x)) / (sin x + sin(2x))

This article explains how to calculate the limit of (2e^x - sqrt(cos x)) / (sin x + sin(2x)) as x approaches 0 using Taylor series expansions and key calculus concepts. It includes a detailed step-by-step solution.

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