Question #a5dc1

1 Answer
Nov 24, 2016

Explanation:

Prove: lim_(xto0)(sin(6x) + sin(8x))/(sin(4x) + sin(6x)) = 7/5

The expression evaluated at the limit is the indeterminate form 0/0, therefore, one should use L'Hôpital's rule

Take the derivative of the numerator:

(d{sin(6x) + sin(8x)})/dx = 6cos(6x) + 8cos(8x)

Take the derivative of the denominator:

(d{sin(4x) + sin(6x)})/dx = 4cos(4x) + 6cos(6x)

Make a new fraction with the new numerator and denominator:

lim_(xto0)(6cos(6x) + 8cos(8x))/(4cos(4x) + 6cos(6x))

The above can be evaluated at the limit:

(6cos(6(0)) + 8cos(8(0)))/(4cos(4(0)) + 6cos(6(0))) = 14/10 = 7/5

The rules says that this is the limit of the original expression:

lim_(xto0)(sin(6x) + sin(8x))/(sin(4x) + sin(6x)) = 7/5

Q.E.D.