2. $(10 * 3+15=45 \mathrm{pts})$ Evaluate the following limit.
(c) $\lim _{x \rightarrow 2} \frac{x-\sqrt{x+2}}{x-2}$
(d) $\lim _{x \rightarrow 1}\left[(x-1)^{2} \sin \frac{1}{x-1}\right]$

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To evaluate the limit $ \lim_{x \rightarrow 2} \frac{x - \sqrt{x + 2}}{x - 2} $, we can substitute $ x = 2 $ directly, which gives us $ \frac{2 - \sqrt{4}}{0} = \frac{0}{0} $, an indeterminate form. Thus, we need to simplify the expression. ‖ ‖ step 2 ⋮ We can rewrite the limit as $ \lim_{x \rightarrow 2} \frac{x - \sqrt{x + 2}}{x - 2} = \frac{-(\sqrt{x + 2} - x)}{x - 2} $. Now, we can factor the numerator. ‖ ‖ step 3 ⋮ Using the identity $ a^2 - b^2 = (a - b)(a + b) $, we can express $ \sqrt{x + 2} - x $ as $ \frac{(x + 2) - x^2}{\sqrt{x + 2} + x} = \frac{2 - (x - 2)(x + 2)}{\sqrt{x + 2} + x} $. Thus, we can simplify further. ‖ ‖ step 4 ⋮ After simplification, we find that $ \lim_{x \rightarrow 2} \frac{-(\sqrt{x + 2} - x)}{x - 2} = -\frac{1}{\sqrt{4} + 2} = -\frac{1}{4} $. Therefore, the limit evaluates to $ -\frac{1}{4} $. ‖ ‖ step 5 ⋮ Now, for the second limit $ \lim_{x \rightarrow 1} (x - 1)^2 \sin\left(\frac{1}{x - 1}\right) $, we notice that as $ x $ approaches 1, $ (x - 1)^2 $ approaches 0. The sine function oscillates between -1 and 1, but since it is multiplied by $ (x - 1)^2 $, the whole expression approaches 0. ‖ ‖ step 6 ⋮ Therefore, we conclude that $ \lim_{x \rightarrow 1} (x - 1)^2 \sin\left(\frac{1}{x - 1}\right) = 0 $. ‖ ∻Answer∻ ⚹ The limit $ \lim_{x \rightarrow 2} \frac{x - \sqrt{x + 2}}{x - 2} $ evaluates to $ -\frac{1}{4} $ and $ \lim_{x \rightarrow 1} (x - 1)^2 \sin\left(\frac{1}{x - 1}\right) $ evaluates to $ 0 $. ⚹ ∻Key Concept∻ ⚹ Limits and indeterminate forms are crucial in calculus for evaluating expressions that do not yield a straightforward result. ⚹ ∻Explanation∻ ⚹ The first limit required simplification due to the indeterminate form, while the second limit approached zero due to the behavior of the sine function multiplied by a term that approaches zero. ⚹◊What technique can be used to evaluate the limit $\lim_{x \to 1} (x-1)^2 \sin\left(\frac{1}{x-1}\right)$?⍭ Generate me a similar question◊

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