If x is so large that terms containing x^{-3} | vedclass

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(B) Given expression: $E = \left(\frac{3x-5}{4x^2+3}\right)^{-4/5} = \left(\frac{4x^2+3}{3x-5}\right)^{4/5} = \left(\frac{4x^2(1+3/(4x^2))}{3x(1-5/(3x

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$\left(\frac{4 x}{3}\right)^{4 / 5}\left(1+\frac{4}{3 x}+\frac{13}{5 x^2}\right)$

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(B) Given expression: $E = \left(\frac{3x-5}{4x^2+3}\right)^{-4/5} = \left(\frac{4x^2+3}{3x-5}\right)^{4/5} = \left(\frac{4x^2(1+3/(4x^2))}{3x(1-5/(3x))}\right)^{4/5} = \left(\frac{4x}{3}\right)^{4/5} \left(1+\frac{3}{4x^2}\right)^{4/5} \left(1-\frac{5}{3x}\right)^{-4/5}$.Using binomial expansion $(1+z)^n \approx 1+nz + \frac{n(n-1)}{2}z^2$:$(1+\frac{3}{4x^2})^{4/5} \approx 1 + \frac{4}{5}(\frac{3}{4x^2}) = 1 + \frac{3}{5x^2}$.$(1-\frac{5}{3x})^{-4/5} \approx 1 + (-\frac{4}{5})(-\frac{5}{3x}) + \frac{(-4/5)(-9/5)}{2}(-\frac{5}{3x})^2 = 1 + \frac{4}{3x} + \frac{18}{25} \cdot \frac{25}{9x^2} = 1 + \frac{4}{3x} + \frac{2}{x^2}$.Multiplying these: $(1 + \frac{3}{5x^2})(1 + \frac{4}{3x} + \frac{2}{x^2}) \approx 1 + \frac{4}{3x} + \frac{2}{x^2} + \frac{3}{5x^2} = 1 + \frac{4}{3x} + \frac{13}{5x^2}$.Thus,$E \approx \left(\frac{4x}{3}\right)^{4/5} \left(1 + \frac{4}{3x} + \frac{13}{5x^2}\right)$.

If $x$ is so large that terms containing $x^{-3}, x^{-4}, x^{-5}, \ldots$ can be neglected,then the approximate value of $\left(\frac{3 x-5}{4 x^2+3}\right)^{-4 / 5}$ is

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