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

Algebra ->  Equations -> 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      Log On


   



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) About Me  (Show Source):
You can put this solution on YOUR website!
What they did was multiply the fraction sin%287x%29%2Fx by 7%2F7, 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 sin%28t%29%2Ft where you replace the 't' with '7x'

All of this is done because normally the limit as x --> 0 has sin%287x%29%2Fx 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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