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We aim to show that 

\[
\sqrt{x^2 + 4} - \sqrt{x^2 + 1} = 1 - \frac{x^2}{4} + \frac{7x^4}{64}
\]

when \(x\) is small. Here's how we proceed:

---

### Step 1: Expand the square roots
We use the binomial expansion for \(\sqrt{1 + u}\) when \(u\) is small:
\[
\sqrt{1 + u} \approx 1 + \frac{u}{2} - \frac{u^2}{8} + \cdots
\]

Rewrite the terms inside the square roots:
1. \(\sqrt{x^2 + 4} = \sqrt{4(1 + \frac{x^2}{4})}\)
\[
\sqrt{x^2 + 4} \approx 2 \left(1 + \frac{x^2}{8} - \frac{x^4}{128}\right) = 2 + \frac{x^2}{4} - \frac{x^4}{64}
\]

2. \(\sqrt{x^2 + 1} = \sqrt{1(1 + x^2)}\)
\[
\sqrt{x^2 + 1} \approx 1 + \frac{x^2}{2} - \frac{x^4}{8}
\]

---

### Step 2: Subtract the expanded terms
Now subtract \(\sqrt{x^2 + 4} - \sqrt{x^2 + 1}\):
\[
\sqrt{x^2 + 4} - \sqrt{x^2 + 1} = \left(2 + \frac{x^2}{4} - \frac{x^4}{64}\right) - \left(1 + \frac{x^2}{2} - \frac{x^4}{8}\right)
\]

Simplify the terms:
1. Constant term: \(2 - 1 = 1\)
2. Coefficient of \(x^2\): \(\frac{x^2}{4} - \frac{x^2}{2} = -\frac{x^2}{4}\)
3. Coefficient of \(x^4\): \(-\frac{x^4}{64} + \frac{x^4}{8} = \frac{7x^4}{64}\)

Thus:
\[
\sqrt{x^2 + 4} - \sqrt{x^2 + 1} = 1 - \frac{x^2}{4} + \frac{7x^4}{64}
\]

---

### Final Answer:
\[
\sqrt{x^2 + 4} - \sqrt{x^2 + 1} = 1 - \frac{x^2}{4} + \frac{7x^4}{64}
\]

---

Do you want further clarification or additional examples?

### 5 Related Questions:
1. What is the general binomial expansion for \((1 + u)^n\) when \(u\) is small?
2. How can this approach be extended to higher-order terms for greater precision?
3. What assumptions are made when \(x\) is considered "small" in this context?
4. Can the same method be applied for differences of cubic roots?
5. How do you verify the accuracy of the expansion numerically?

### Tip:
When dealing with approximations, always check the order of terms retained to ensure consistency in precision.

When x is small, show that sqrt(x^2 + 4) - sqrt(x^2 + 1) = 1 - x^2/4 + 7x^4/64.

Explore the step-by-step derivation for approximating sqrt(x^2 + 4) - sqrt(x^2 + 1) using the binomial expansion and algebra. Suitable for advanced high school or college students.

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