Find a two-term asymptotic expansion, for $\varepsilon \to 0$, of each solution $x$ of \begin{equation} x^{2+\varepsilon} = \frac{1}{x+2\varepsilon},\qquad (x > 0) \end{equation} My approach: First rewriting the equation we get: \begin{align} x^{2+\varepsilon}(x+2\varepsilon)-1 &= 0,\qquad (x > 0) \\ x^{3+\varepsilon}+2\varepsilon x^{2+\varepsilon}-1 &= 0,\qquad (x>0) \end{align} Using $x = \mu X(\varepsilon)$, dominant balance and find $\mu = \varepsilon$. This yields \begin{equation} \varepsilon^{3+\varepsilon}X^{3+\varepsilon}+2\varepsilon^{3+\varepsilon}X^{2+\varepsilon}-1 = 0 \end{equation} From this point I don't know how to proceed.
Two-term asymptotic approximation for roots of an equation
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asymptotics
roots
perturbation-theory