3. Evaluate the following limits using any method we've discussed so far in the course:
(a) $\lim _{x \rightarrow 0} \frac{5 \sin (6 x) \sin (2 x)}{9 x^{2}}$
(b) $\lim _{x \rightarrow \infty}\left(\sqrt{4 x^{4}+9 x^{2}}-2 x^{2}\right)$

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To evaluate the limit as $ x $ approaches 0 of $ \frac{5 \sin(6x) \sin(2x)}{9x^2} $, we can use the Taylor series expansion for $ \sin(x) $: $ \sin(x) \approx x $ near $ x = 0 $. Thus, we have: ‖ ‖ step 2 ⋮ Substituting the Taylor series, we get $ \sin(6x) \approx 6x $ and $ \sin(2x) \approx 2x $. Therefore, the limit becomes: ‖ ‖ step 3 ⋮ $ \lim_{x \rightarrow 0} \frac{5(6x)(2x)}{9x^2} = \lim_{x \rightarrow 0} \frac{60x^2}{9x^2} = \frac{60}{9} = \frac{20}{3} $. ‖ ‖ step 4 ⋮ The final result for part (a) is $ \frac{20}{3} $. ‖ ‖ step 5 ⋮ For part (b), we evaluate the limit as $ x $ approaches infinity of $ \sqrt{4x^4 + 9x^2} - 2x^2 $. We can factor out $ x^4 $ from the square root: ‖ ‖ step 6 ⋮ This gives us $ \lim_{x \rightarrow \infty} \left( \sqrt{x^4(4 + \frac{9}{x^2})} - 2x^2 \right) = \lim_{x \rightarrow \infty} \left( x^2\sqrt{4 + \frac{9}{x^2}} - 2x^2 \right) $. ‖ ‖ step 7 ⋮ Simplifying further, we have $ x^2(\sqrt{4 + \frac{9}{x^2}} - 2) $. As $ x $ approaches infinity, $ \sqrt{4 + \frac{9}{x^2}} $ approaches $ 2 $, leading to: ‖ ‖ step 8 ⋮ $ \lim_{x \rightarrow \infty} x^2(2 - 2) = 0 $. Thus, we need to consider the next term in the expansion, which gives us $ \frac{9}{4} $ as the limit. ‖ ∻Answer∻ ⚹ Part (a): $ \frac{20}{3} $; Part (b): $ \frac{9}{4} $ ⚹ ∻Key Concept∻ ⚹ Limits and Taylor series expansion are used to evaluate limits at points where direct substitution leads to indeterminate forms. ⚹ ∻Explanation∻ ⚹ The Taylor series allows us to approximate functions near a point, making it easier to evaluate limits that would otherwise be difficult to compute directly. In this case, we used it to find the limits as $ x $ approaches 0 and infinity. ⚹◊How can you apply L'Hôpital's Rule to evaluate the limit $\lim_{x \rightarrow 0} \frac{5 \sin(6x) \sin(2x)}{9x^2}$?⍭ Generate me a similar question◊

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