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Question Number 175880 by cortano1 last updated on 08/Sep/22

Answered by blackmamba last updated on 09/Sep/22

$$\underset{x\to 0}{\lim }\frac{{bx}^2{\left(\frac{\sin x}{x}\right)}^2+3x\left(\frac{\tan 3x}{3x}\right)+2x^2{\left(\frac{\sin \frac{1}{2}x}{x}\right)}^2}{\sqrt{2}x\left(\frac{\sin \frac{3}{2}x}{x}\right)+x^{2b}{\left(\frac{\sin 3x}{x}\right)}^{2b}}$$$$=\underset{x\to 0}{\lim }\left[\frac{{bx}^2+3x+\frac{1}{2}{x}^2}{\frac{3\sqrt{2}}{2}x+{\left(9x\right)}^{2b}}\right]$$$$=\underset{x\to 0}{\lim }\left[\frac{bx+3+\frac{1}{2}x}{\frac{3\sqrt{2}}{2}+9^{2b}.x^{2b-1}}\right]$$$$=\frac{2.3}{3\sqrt{2}}=\sqrt{2}$$

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