5. Show that \( \lim _{x \rightarrow 9} \sqrt{x}=3 \).
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To demonstrate that \( \lim_{x \rightarrow 9} \sqrt{x} = 3 \), we need to consider the definition of a limit. As \( x \) approaches 9, we look at the values of \( \sqrt{x} \). When \( x = 9 \), then \( \sqrt{9} = 3 \). To make this more rigorous, for any \( \epsilon > 0 \), we want to find a \( \delta > 0 \) such that whenever \( 0 < |x - 9| < \delta \), it follows that \( |\sqrt{x} - 3| < \epsilon \). Using the inequality \( |\sqrt{x} - 3| < \epsilon \), we can manipulate it as such: \[ -\epsilon < \sqrt{x} - 3 < \epsilon \] This can be rewritten as: \[ 3 - \epsilon < \sqrt{x} < 3 + \epsilon \] Squaring all parts (assuming \( \epsilon < 3 \) to maintain positive values under the square root), we get: \[ (3 - \epsilon)^2 < x < (3 + \epsilon)^2 \] Calculating the bounds gives: \[ 9 - 6\epsilon + \epsilon^2 < x < 9 + 6\epsilon + \epsilon^2 \] Thus, we can choose \( \delta \) to be the minimum of \( 6\epsilon \) or \( \epsilon^2 \): Set \( \delta = \min(6\epsilon, \epsilon^2) \). This ensures that \( |x - 9| < \delta \) will indeed keep \( |\sqrt{x} - 3| < \epsilon \). Since we can make \( |\sqrt{x} - 3| < \epsilon \) as \( x \) approaches 9, we conclude that: \[ \lim_{x \rightarrow 9} \sqrt{x} = 3. \]
