SOLUTION: Please explain this solution: lim(x→0) sin(7x) cos(3x)/x = lim(x→0) cos(3x) * sin(7x)/x = lim(x→0) cos(3x) * 7sin(7x)/(7x) = cos 0 * (7 * 1), since lim(t

Question 987818: Please explain this solution:
lim(x→0) sin(7x) cos(3x)/x
= lim(x→0) cos(3x) * sin(7x)/x
= lim(x→0) cos(3x) * 7sin(7x)/(7x)
= cos 0 * (7 * 1), since lim(t→0) sin(t)/t = 1 with t = 7x
= 7.

What causes cos(3x)*sin(7x)/x to turn into cos(3x)*7sin(7x)/(7x)? It looks like it is a common denominator of a fraction with sin but wouldn't that be invalid since it's sin(7x)/x isn't the sin(7x) internally connected and therefore cannot be broken into some like 7sin(7x)/7x. Or can it be thought of like sin/x*sin(7x)? Also, why doesn't cos(3x)/x become 3cos(3x)/3x or both of them together. Why is one chosen and the other gets left alone?

Thanks

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
What they did was multiply the fraction by , which is a fancy form of 1. Multiplying by this form of 1 does not change the expression

The reason why they did this was they wanted it in the form where you replace the 't' with '7x'

All of this is done because normally the limit as x --> 0 has potentially undefined (because of that division by zero). There are no errors when you plug in x = 0 into the cos(3x) portion, so no algebraic tricks are needed there.

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If you need more help, or if you have any questions about the problem, feel free to email me at jim_thompson5910@hotmail.com
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SOLUTION: Please explain this solution: lim(x&#8594;0) sin(7x) cos(3x)/x = lim(x&#8594;0) cos(3x) * sin(7x)/x = lim(x&#8594;0) cos(3x) * 7sin(7x)/(7x) = cos 0 * (7 * 1), since lim(t

SOLUTION: Please explain this solution: lim(x→0) sin(7x) cos(3x)/x = lim(x→0) cos(3x) * sin(7x)/x = lim(x→0) cos(3x) * 7sin(7x)/(7x) = cos 0 * (7 * 1), since lim(t