The question in the image reads:

"If x be so small that its square and higher powers may be neglected, then evaluate" the following expression:

$\frac{(1+x)^n}{(9+2x)^{1/2}}$

To solve this, let's apply the approximation method that takes advantage of the fact that $x$ is very small. This means we can neglect the terms involving $x^2$ and higher powers of $x$.

Step 1: Approximate the expression for small $x$

We start by approximating both the numerator and denominator for small $x$.

Numerator: $(1+x)^n$

For small $x$, we can use the binomial approximation: (1+x)^n \approx 1 + nx \quad \text{(neglecting higher powers of x)}

Denominator: $(9+2x)^{1/2}$

We can expand $(9+2x)^{1/2}$ using the binomial approximation for fractional powers: $(9+2x)^{1/2} = 9^{1/2} \left( 1 + \frac{2x}{9} \right)^{1/2}$ For small $x$, we approximate this as: $(9+2x)^{1/2} \approx 3 \left( 1 + \frac{x}{9} \right) \approx 3 + \frac{x}{3}$

Step 2: Putting it together

Now substitute the approximations into the original expression: $\frac{(1+x)^n}{(9+2x)^{1/2}} \approx \frac{1 + nx}{3 + \frac{x}{3}}$

Step 3: Simplify the expression

To simplify the expression, we can use the approximation $3 + \frac{x}{3} \approx 3$ for small $x$, which leads to: $\frac{1 + nx}{3} = \frac{1}{3} + \frac{nx}{3}$

Thus, the simplified result is approximately: $\frac{1}{3} + \frac{nx}{3}$

Final Answer:

The approximate value of the expression is: $\frac{1}{3} + \frac{nx}{3}$

Would you like further clarification or any details on this solution? Here are a few related questions:

1. What is the binomial approximation for $(1+x)^n$ when $x$ is small?
2. How do you apply the binomial expansion for fractional powers like $(9+2x)^{1/2}$?
3. How can this approximation method be generalized for other types of expressions involving small variables?
4. What happens if higher-order terms of $x$ are not neglected? How would the result change?
5. What is the role of neglecting higher powers of $x$ in simplifying expressions?

Tip: When approximating for small values of $x$, it is crucial to only keep the first few terms of the expansion and discard higher-order terms to make the calculation manageable and quick.