Question #2880c

1 Answer
Dec 14, 2017

Multiply the top and bottom by the fraction (60x)/(60x) before rearranging and using the given limit to obtain an answer of 20/3.

Explanation:

Let f(x)=(sin(5x)cot(3x))/(xcot(4x)). Then

f(x)=(60x)/(60x)*f(x)=(60xsin(5x)cot(3x))/(60x^{2}cot(4x))

=(60xsin(5x)cos(3x)sin(4x))/(60x^{2}sin(3x)cos(4x))

=(3x)/(sin(3x))* sin(5x)/(5x) * sin(4x)/(4x) * (20cos(3x))/(3cos(4x))

Since lim_{x->0}sin(x)/x=1, it follows that lim_{x->0}(3x)/(sin(3x))=1, lim_{x->0}sin(5x)/(5x)=1, and lim_{x->0}sin(4x)/(4x)=1.

Moreover, we can say that lim_{x->0}(20cos(3x))/(3cos(4x))=(20 * 1)/(3 *1) = 20/3 by substitution into the function since (20cos(3x))/(3cos(4x)) is continuous at x=0.

Finally, applying the rule for the limit of a product of functions (whose individual limits exist), we obtain

lim_{x->0}f(x)=1 * 1 * 1 * 20/3 = 20/3.