When x is so small that its square and its hi | vedclass

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Question

(A) Given that $x^n \approx 0$ for $n \ge 2$. Let $y = \frac{(1+\frac{3}{4}x)^{-4} \sqrt{3+x}}{\sqrt{(3-x)^3}}$. $y = \frac{(1+\frac{3}{4}x)^{-4} \sqr

Vedclass · Accepted Answer

$\frac{1}{3}-\frac{7 x}{9}$

Vedclass · Answer

(A) Given that $x^n \approx 0$ for $n \ge 2$. Let $y = \frac{(1+\frac{3}{4}x)^{-4} \sqrt{3+x}}{\sqrt{(3-x)^3}}$. $y = \frac{(1+\frac{3}{4}x)^{-4} \sqrt{3}(1+\frac{x}{3})^{1/2}}{3^{3/2}(1-\frac{x}{3})^{3/2}}$. $y = \frac{1}{3} (1+\frac{3}{4}x)^{-4} (1+\frac{x}{3})^{1/2} (1-\frac{x}{3})^{-3/2}$. Using the binomial expansion $(1+z)^n \approx 1+nz$ for small $z$: $y \approx \frac{1}{3} (1 - 4 \cdot \frac{3}{4}x) (1 + \frac{1}{2} \cdot \frac{x}{3}) (1 - (-\frac{3}{2}) \cdot \frac{x}{3})$. $y \approx \frac{1}{3} (1 - 3x) (1 + \frac{x}{6}) (1 + \frac{x}{2})$. $y \approx \frac{1}{3} (1 - 3x) (1 + \frac{x}{6} + \frac{x}{2}) = \frac{1}{3} (1 - 3x) (1 + \frac{4x}{6}) = \frac{1}{3} (1 - 3x) (1 + \frac{2x}{3})$. $y \approx \frac{1}{3} (1 + \frac{2x}{3} - 3x - 2x^2) \approx \frac{1}{3} (1 - \frac{7x}{3})$. $y \approx \frac{1}{3} - \frac{7x}{9}$.

When $x$ is so small that its square and its higher powers may be neglected,then the value of $\frac{\left(1+\frac{3}{4} x\right)^{-4} \sqrt{(3+x)}}{\sqrt{(3-x)^3}}$ is approximately equal to

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