lim x->0 of sin(2x)sin(7x)/xsin(9x) evaluate the limit. please help?
combine the x in numerator with either sin2x or sin 7x then multiply and divide by x
i would go with my first guess of \(\frac{2\times 7}{9}\) but i suppose that is not a good explanation
\[\lim_{x \to 0}\frac{\sin(ax)}{\sin(bx)}=\frac{a}{b}\]
do you have to do each limit individually?
damn typo
\[\lim_{x\to 0}\frac{\sin(2x)}{\sin(9x)}\times \lim_{x\to 0}\frac{\sin(7x)}{x}\]
you can use the product of the limits
why are you multiplying the denominator with the numerator?
hope it is clear that the first one is \[\frac{2}{9}\] and the second is 7
i am not unless i made another mistake
did you mix up the sin7x with the sin9x?
\[\frac{\sin(2x)\sin(7x)}{x\sin(9x)}\] is what you start with right?
correct : )
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
i don't understand why the sin(9x) is now under the sin(2x) instead of sin(7x)
either way right?
\[\frac{ab}{cd}=\frac{a}{c}\times \frac{b}{d}=\frac{a}{d}\times \frac{b}{c}\]
pick one it makes no difference which one you choose to use
so now i multiply the first limit by 2/9 and the second limit by 7/7?
the idea is this: \[\lim_{x\to 0}\frac{\sin(ax)}{x}=a\] and \[\lim_{x\to 0}\frac{\sin(bx)}{\sin(cx)}=\frac{b}{c}\]
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
right, so i should end up with lim x-> 0 of \[\frac{ \sin2x }{ \sin9x }\times \frac{ 2 }{ 9 }+\frac{ \sin7x }{ x }\times \frac{ 7 }{ 7 }\]
right but you need a times sign not a plus sign
bring the 2/9 infront of the limit sign, same with the 7 and end up with 2/9times7?
so it should be 2/9t x 7?
yes, without the \(t\)
answer being 14/9?
yes
thank you very much for your help! : )
yw
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