Answered

At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

[tex]\lim_{x \to 0 } \frac{x-sin(2x)}{x+sin(3x)}[/tex]

Sagot :

syahbanazacki

Step-by-step explanation:

[tex] = \lim \limits_{x \to0} \frac{x - \sin(2x) }{x + \sin(3x) } [/tex]

[tex] = \lim \limits_{x \to0} \frac{ \frac{d}{dx}(x - \sin(2x) ) }{ \frac{d}{dx} (x + \sin(3x) )} [/tex]

[tex] = \lim \limits_{x \to0} \frac{1 - 2\cos(2x) }{1 + 3 \cos(3x) } [/tex]

[tex] = \lim \limits_{x \to0} \frac{ \frac{d}{dx}(1 + 2 \cos(2x) ) }{ \frac{d}{dx} (1 + 3 \cos(3x) )} [/tex]

[tex] = \lim \limits_{x \to0} \frac{ - 4 \sin(2x) }{ - 9 \sin(3x) } [/tex]

[tex] = \frac{ - 4.2}{ - 9.3} [/tex]

[tex] = \frac{ - 8}{ - 27} [/tex]

[tex] = \frac{8}{27} [/tex]

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.