Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
[tex]\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}[/tex]
Step-by-step explanation:
Notice that [tex]\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=\frac{12(0)e^{0}-12(0)}{cos(5(0))-1}=\frac{0}{0}[/tex], which is in indeterminate form, so we must use L'Hôpital's rule which states that [tex]\lim_{x \to c} \frac{f(x)}{g(x)}=\lim_{x \to c} \frac{f'(x)}{g'(x)}[/tex]. Basically, we keep differentiating the numerator and denominator until we can plug the limit in without having any discontinuities:
[tex]\frac{12xe^x-12x}{cos(5x)-1}\\\\\frac{12xe^x+12e^x-12}{-5sin(5x)}\\ \\\frac{12xe^x+12e^x+12e^x}{-25cos(5x)}[/tex]
Now, plug in the limit and evaluate:
[tex]\frac{12(0)e^{0}+12e^{0}+12e^{0}}{-25cos(5(0))}\\ \\\frac{12+12}{-25cos(0)}\\ \\\frac{24}{-25}\\ \\-\frac{24}{25}[/tex]
Thus, [tex]\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}[/tex]
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
