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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]
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