\lim \:_{x\to \:-\infty \:}\left(\arcsin \left(\frac{x}{\sqrt{4x^2+1}}\right)\right)

To solve the limit \(\lim_{x \to -\infty} \left(\arcsin \left(\frac{x}{\sqrt{4x^2+1}}\right)\right)\), let's follow these steps:

1. **Simplify the expression inside the arcsine:**

   Consider the expression inside the arcsine:
   \[
   \frac{x}{\sqrt{4x^2 + 1}}
   \]

   As \(x \to -\infty\), \(x^2\) becomes very large. So, the term \(1\) in the denominator becomes negligible compared to \(4x^2\). Thus, the expression can be approximated as:
   \[
   \frac{x}{\sqrt{4x^2 + 1}} \approx \frac{x}{\sqrt{4x^2}} = \frac{x}{\sqrt{4} \cdot \sqrt{x^2}} = \frac{x}{2|x|}
   \]

   Since \(x\) is approaching \(-\infty\), \( |x| = -x \). Therefore:
   \[
   \frac{x}{2|x|} = \frac{x}{2(-x)} = -\frac{1}{2}
   \]

2. **Apply the limit:**

   Now that we have:
   \[
   \lim_{x \to -\infty} \frac{x}{\sqrt{4x^2 + 1}} = -\frac{1}{2}
   \]

   Substitute this result into the original limit:
   \[
   \lim_{x \to -\infty} \arcsin \left(\frac{x}{\sqrt{4x^2 + 1}}\right) = \arcsin \left(-\frac{1}{2}\right)
   \]

3. **Evaluate the arcsine function:**

   The arcsine of \(-\frac{1}{2}\) is:
   \[
   \arcsin \left(-\frac{1}{2}\right) = -\frac{\pi}{6}
   \]

Thus, the limit is:
\[
\lim_{x \to -\infty} \left(\arcsin \left(\frac{x}{\sqrt{4x^2+1}}\right)\right) = -\frac{\pi}{6}
\]

Would you like further details or have any questions?

Here are five related questions for further practice:
1. What is \(\lim_{x \to \infty} \left(\arcsin \left(\frac{x}{\sqrt{4x^2+1}}\right)\right)\)?
2. How do you calculate \(\lim_{x \to -\infty} \frac{x}{\sqrt{x^2 + 1}}\)?
3. What is the value of \(\lim_{x \to \infty} \arctan(x)\)?
4. Can you compute \(\lim_{x \to -\infty} \left(\arccos \left(\frac{x}{\sqrt{4x^2+1}}\right)\right)\)?
5. How would you find \(\lim_{x \to \infty} \frac{\sqrt{x^2+1} - \sqrt{x^2-1}}{x}\)?

**Tip:** When evaluating limits involving inverse trigonometric functions, it's often helpful to first simplify the expression inside the function, especially as the variable approaches infinity.

Learn how to evaluate the limit of arcsin(x / sqrt(4x^2 + 1)) as x approaches negative infinity with detailed steps and solution. This problem involves limits and inverse trigonometric functions, suitable for advanced high school students and college-level mathematics.

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