.6578 this would be the answer
The answer to this question is -5
Answer:
![\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B12xe%5Ex-12x%7D%7Bcos%285x%29-1%7D%3D-%5Cfrac%7B24%7D%7B25%7D)
Step-by-step explanation:
Notice that
, which is in indeterminate form, so we must use L'Hôpital's rule which states that
. Basically, we keep differentiating the numerator and denominator until we can plug the limit in without having any discontinuities:
![\frac{12xe^x-12x}{cos(5x)-1}\\\\\frac{12xe^x+12e^x-12}{-5sin(5x)}\\ \\\frac{12xe^x+12e^x+12e^x}{-25cos(5x)}](https://tex.z-dn.net/?f=%5Cfrac%7B12xe%5Ex-12x%7D%7Bcos%285x%29-1%7D%5C%5C%5C%5C%5Cfrac%7B12xe%5Ex%2B12e%5Ex-12%7D%7B-5sin%285x%29%7D%5C%5C%20%5C%5C%5Cfrac%7B12xe%5Ex%2B12e%5Ex%2B12e%5Ex%7D%7B-25cos%285x%29%7D)
Now, plug in the limit and evaluate:
![\frac{12(0)e^{0}+12e^{0}+12e^{0}}{-25cos(5(0))}\\ \\\frac{12+12}{-25cos(0)}\\ \\\frac{24}{-25}\\ \\-\frac{24}{25}](https://tex.z-dn.net/?f=%5Cfrac%7B12%280%29e%5E%7B0%7D%2B12e%5E%7B0%7D%2B12e%5E%7B0%7D%7D%7B-25cos%285%280%29%29%7D%5C%5C%20%5C%5C%5Cfrac%7B12%2B12%7D%7B-25cos%280%29%7D%5C%5C%20%5C%5C%5Cfrac%7B24%7D%7B-25%7D%5C%5C%20%5C%5C-%5Cfrac%7B24%7D%7B25%7D)
Thus, ![\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B12xe%5Ex-12x%7D%7Bcos%285x%29-1%7D%3D-%5Cfrac%7B24%7D%7B25%7D)
Step-by-step explanation:
use the formula if things get to difficult