In the following questions find the limit \( \begin{array}{ll}\text { 17. a. } \lim _{x \rightarrow-2^{+}}(x+3) \frac{|x+2|}{x+2} & \text { b. } \lim _{x \rightarrow-2^{-}}(x+3) \frac{|x+2|}{x+2} \\ \text { 18. a. } \lim _{x \rightarrow 1^{+}} \frac{\sqrt{2 x}(x-1)}{|x-1|} & \text { b. } \lim _{x \rightarrow 1^{-}} \frac{\sqrt{2 x}(x-1)}{|x-1|}\end{array} \)
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Let's dive into these limits with a bit of flair! For the first limits, when \( x \rightarrow -2^{+}\) in part a, we see that \( |x + 2| = x + 2 \) since \( x + 2 \) is positive just to the right of -2. Therefore, we have: \[ \lim _{x \rightarrow -2^{+}}(x + 3) \frac{|x + 2|}{x + 2} = \lim _{x \rightarrow -2^{+}}(x + 3) = -2 + 3 = 1. \] For part b, as \( x \rightarrow -2^{-}\), the expression \( |x + 2| = -(x + 2) \) since it's negative just to the left of -2. Then: \[ \lim _{x \rightarrow -2^{-}}(x + 3) \frac{|x + 2|}{x + 2} = \lim _{x \rightarrow -2^{-}}(x + 3)(-1) = -(-2 + 3) = -1. \] Now onto the next limits! For part a, as \( x \rightarrow 1^{+}\), we need to be cautious with the absolute value in the denominator: \[ \lim _{x \rightarrow 1^{+}} \frac{\sqrt{2x}(x - 1)}{|x - 1|} = \lim _{x \rightarrow 1^{+}} \sqrt{2x} \cdot \frac{x - 1}{x - 1} = \lim _{x \rightarrow 1^{+}} \sqrt{2x} = \sqrt{2 \cdot 1} = \sqrt{2}. \] For part b, as \( x \rightarrow 1^{-}\): \[ \lim _{x \rightarrow 1^{-}} \frac{\sqrt{2x}(x - 1)}{|x - 1|} = \lim _{x \rightarrow 1^{-}} \sqrt{2x} \cdot \frac{x - 1}{-(x - 1)} = -\lim _{x \rightarrow 1^{-}} \sqrt{2x} = -\sqrt{2}. \] Spicing it up, we found that these limits reveal how the behavior of functions can shift dramatically based on their approach toward critical points! 📈 Enjoy exploring these limits!
