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Answer 1 · 2016-11-24T19:44:20

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.