the limit \( \lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}=2 \), illustrate this definition by finding the largest possible values of \( \delta \) that correspond to \( \varepsilon=0.4 \) and \( \varepsilon=0.2 \). (Round your answers to three decimal places. \( =0.4 \quad \delta=\square \) \( =0.2 \quad \delta=\square \) Jeed Help? Raadit Ubmit Answer

To find the largest possible values of \( \delta \) that correspond to \( \varepsilon = 0.4 \) and \( \varepsilon = 0.2 \) for the limit \[ \lim_{x \rightarrow 0} \frac{e^{2x} - 1}{x} = 2, \] we can use the definition of the limit. According to this definition, for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x| < \delta \), then \[ \left| \frac{e^{2x} - 1}{x} - 2 \right| < \varepsilon. \] ### Step 1: Simplifying the expression We start with the expression: \[ \left| \frac{e^{2x} - 1}{x} - 2 \right| < \varepsilon. \] This can be rewritten as: \[ \left| \frac{e^{2x} - 1 - 2x}{x} \right| < \varepsilon. \] ### Step 2: Using Taylor expansion Using the Taylor expansion of \( e^{2x} \) around \( x = 0 \): \[ e^{2x} = 1 + 2x + \frac{(2x)^2}{2} + O(x^3) = 1 + 2x + 2x^2 + O(x^3). \] Thus, \[ e^{2x} - 1 = 2x + 2x^2 + O(x^3). \] Substituting this back into our limit expression gives: \[ \frac{e^{2x} - 1}{x} = 2 + 2x + O(x^2). \] So, \[ \frac{e^{2x} - 1}{x} - 2 = 2x + O(x^2). \] ### Step 3: Finding \( \delta \) Now we need to find \( \delta \) such that: \[ \left| 2x + O(x^2) \right| < \varepsilon. \] For small \( x \), \( O(x^2) \) is negligible compared to \( 2x \). Thus, we can approximate: \[ |2x| < \varepsilon. \] This leads to: \[ |x| < \frac{\varepsilon}{2}. \] ### Step 4: Calculate \( \delta \) for given \( \varepsilon \) 1. For \( \varepsilon = 0.4 \): \[ \delta < \frac{0.4}{2} = 0.2. \] 2. For \( \varepsilon = 0.2 \): \[ \delta < \frac{0.2}{2} = 0.1. \] ### Final Answers Thus, rounding to three decimal places, we have: \[ \varepsilon = 0.4 \quad \delta = 0.200, \] \[ \varepsilon = 0.2 \quad \delta = 0.100. \]