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Mathematics 16 Online
OpenStudy (anonymous):

lim x->0 of sin(2x)sin(7x)/xsin(9x) evaluate the limit. please help?

hartnn (hartnn):

combine the x in numerator with either sin2x or sin 7x then multiply and divide by x

OpenStudy (anonymous):

i would go with my first guess of \(\frac{2\times 7}{9}\) but i suppose that is not a good explanation

OpenStudy (anonymous):

\[\lim_{x \to 0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b}\]

OpenStudy (anonymous):

do you have to do each limit individually?

OpenStudy (anonymous):

damn typo

OpenStudy (anonymous):

\[\lim_{x\to 0}\frac{\sin(2x)}{\sin(9x)}\times \lim_{x\to 0}\frac{\sin(7x)}{x}\]

OpenStudy (anonymous):

you can use the product of the limits

OpenStudy (anonymous):

why are you multiplying the denominator with the numerator?

OpenStudy (anonymous):

hope it is clear that the first one is \[\frac{2}{9}\] and the second is 7

OpenStudy (anonymous):

i am not unless i made another mistake

OpenStudy (anonymous):

did you mix up the sin7x with the sin9x?

OpenStudy (anonymous):

\[\frac{\sin(2x)\sin(7x)}{x\sin(9x)}\] is what you start with right?

OpenStudy (anonymous):

correct : )

OpenStudy (anonymous):

you can break it up into any product you choose. i chose \[\frac{\sin(2x)}{\sin(9x)}\times \frac{\sin(7x)}{x}\] but you could split it another way if you like

OpenStudy (anonymous):

i don't understand why the sin(9x) is now under the sin(2x) instead of sin(7x)

OpenStudy (anonymous):

either way right?

OpenStudy (anonymous):

\[\frac{ab}{cd}=\frac{a}{c}\times \frac{b}{d}=\frac{a}{d}\times \frac{b}{c}\]

OpenStudy (anonymous):

pick one it makes no difference which one you choose to use

OpenStudy (anonymous):

so now i multiply the first limit by 2/9 and the second limit by 7/7?

OpenStudy (anonymous):

the idea is this: \[\lim_{x\to 0}\frac{\sin(ax)}{x}=a\] and \[\lim_{x\to 0}\frac{\sin(bx)}{\sin(cx)}=\frac{b}{c}\]

OpenStudy (anonymous):

so you are going to get \[\frac{2}{9}\times 7\] or if you do it the other way you are going to get \[2\times \frac{7}{9}\] makes no difference

OpenStudy (anonymous):

right, so i should end up with lim x-> 0 of \[\frac{ \sin2x }{ \sin9x }\times \frac{ 2 }{ 9 }+\frac{ \sin7x }{ x }\times \frac{ 7 }{ 7 }\]

OpenStudy (anonymous):

right but you need a times sign not a plus sign

OpenStudy (anonymous):

bring the 2/9 infront of the limit sign, same with the 7 and end up with 2/9times7?

OpenStudy (anonymous):

so it should be 2/9t x 7?

OpenStudy (anonymous):

yes, without the \(t\)

OpenStudy (anonymous):

answer being 14/9?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

thank you very much for your help! : )

OpenStudy (anonymous):

yw

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