Question

Limit question

Answer 1

Find the lim as x -> 0 for sin(5x)cos(3x) / sin(2x)

@callisto

Answer 2

L'hospitals law?

Answer 3

\[lim_{x\rightarrow 0} \frac{sin(5x)cos(3x)}{sin(2x)}\]\[=lim_{x\rightarrow 0} \frac{sin(5x)}{sin(2x)} \times lim_{x\rightarrow 0}cos(3x)\]\[=lim_{x\rightarrow 0} \frac{\frac{sin(5x)}{10x}}{\frac{sin(2x)}{10x}} \times lim_{x\rightarrow 0}cos(3x)\]\[=\frac{\frac{1}{2}lim_{x\rightarrow 0} \frac{sin(5x)}{5x}}{\frac{1}{5}lim_{x\rightarrow 0} \frac{sin(2x)}{2x}} \times lim_{x\rightarrow 0}cos(3x)\]

Answer 4

Recall:\[lim_{x\rightarrow 0} \frac{sinx}{x} =1\]And\[lim_{x\rightarrow 0} cosx =1\]

Answer 5

why 10x below sin5x and sin2x?

Answer 6

Because... 10x is the LCM of 5x and 2x :|

Answer 7

still dont understand how you deduced an LCM in the above problem specifically, where did that came from? please explain?

Answer 8

are you applying L'Hospitals law or just manipulation?

Answer 9

AH, I understand now. You are manipulating to refer it to recalling sinx/x = 1. In this case our "x" is "5x" hence sin5x/5x = 1 likewise for sin2x/2x = 1.

Answer 10

Yup :)
PS: Sorry, I was away..

Answer 11

just a quick question, how would i determine what to times by when simplifying the LCM? Like you did above with 1/2 and 5/2?

Also just clarify, you were doing L'Hospitals law or just manipulation?

Answer 12

Simplifying LCM? I actually divided both numerator and denominator of the fraction by the LCM

I was doing manipulation only. When you use L'Hopital's rule, you need differentiation.

Answer 13

@Callisto both ways are ok, the same result

Answer 14

I know, but I did it by manipulation only.

Answer 15

yes, you are right. Your way is perfect since it needs flexibility and experience.

Answer 16

actually, u can use this :
lim (x->0) sin(ax)/sin(bx) = a/b