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16. Find the limit if it exists: \( \lim _{x \rightarrow 0}(1+x)^{\frac{1}{2 x}} \) Hint Let \( y=(1+x)^{\frac{1}{2 x}} \). Take Logs

Ask by Murray Welch. in South Africa
Mar 15,2025

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The limit is \( e^{\frac{1}{2}} \).

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To find the limit \( \lim _{x \rightarrow 0}(1+x)^{\frac{1}{2 x}} \), we start by letting \( y=(1+x)^{\frac{1}{2 x}} \) and then take the natural logarithm of both sides: \[ \ln y = \frac{1}{2x} \ln(1+x). \] Now, we need to find the limit as \( x \) approaches 0. Using the Taylor series expansion for \( \ln(1+x) \) around \( x=0 \), we have: \[ \ln(1+x) \approx x \quad \text{for small } x. \] Thus, we can rewrite: \[ \ln y \approx \frac{1}{2x} \cdot x = \frac{1}{2}. \] Now we can evaluate the limit: \[ \lim_{x \rightarrow 0} \ln y = \frac{1}{2}. \] Exponentiating both sides yields: \[ \lim_{x \rightarrow 0} y = e^{\frac{1}{2}} = \sqrt{e}. \] Therefore, the limit is: \[ \lim _{x \rightarrow 0}(1+x)^{\frac{1}{2 x}} = \sqrt{e}. \]

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