Question

Evaluate the following limit, if it exists : limx→0 (12xe^x−12x) / (cos(5x)−1)

Answer 1

Answer:

$\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}$

Step-by-step explanation:

Notice that $\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=\frac{12(0)e^{0}-12(0)}{cos(5(0))-1}=\frac{0}{0}$, which is in indeterminate form, so we must use L'Hôpital's rule which states that $\lim_{x \to c} \frac{f(x)}{g(x)}=\lim_{x \to c} \frac{f'(x)}{g'(x)}$. Basically, we keep differentiating the numerator and denominator until we can plug the limit in without having any discontinuities:

$\frac{12xe^x-12x}{cos(5x)-1}\\\\\frac{12xe^x+12e^x-12}{-5sin(5x)}\\ \\\frac{12xe^x+12e^x+12e^x}{-25cos(5x)}$

Now, plug in the limit and evaluate:

$\frac{12(0)e^{0}+12e^{0}+12e^{0}}{-25cos(5(0))}\\ \\\frac{12+12}{-25cos(0)}\\ \\\frac{24}{-25}\\ \\-\frac{24}{25}$

Thus, $\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}$