okay so i'm confused as to HOW to go about finding the limits of these trigonometric fxns.... lim x-->0 tan(2x) / sin(3x)cos(4x) and lim x-->0 sin(x^2 - 9) / x-3
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Originally Posted by veezey11 okay so i'm confused as to HOW to go about finding the limits of these trigonometric fxns.... lim x-->0 tan(2x) / sin(3x)cos(4x) We should use the fact that and Thus, Does this make sense? --Chris
uhhhm sortof yeah... i'll take a closer look at it. and see if i can figure it all out. THANK YOU SO MUCH! <3
Originally Posted by veezey11 okay so i'm confused as to HOW to go about finding the limits of these trigonometric fxns.... lim x-->0 tan(2x) / sin(3x)cos(4x) and lim x-->0 sin(x^2 - 9) / x-3 The easier way is with L'Hospital's rule, namely . So Can you go from here?
Originally Posted by veezey11 okay so i'm confused as to HOW to go about finding the limits of these trigonometric fxns.... lim x-->0 tan(2x) / sin(3x)cos(4x) and lim x-->0 sin(x^2 - 9) / x-3 For the second, why can't you just use direct substitution?
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}{\sin(3x)\cos(4x)}=\lim_{x\to0}\f rac{\tan(2x)}{\sin(3x)}\cdot\underbrace{\lim_{x\to 0}\frac{1}{\cos(4x)}}_1=\lim_{x\to0}\frac{\display style\frac{\tan(2x)}{x}}{\displaystyle\frac{\sin(3 x)}{x}}=\lim_{x\to0}\frac{\displaystyle\frac{\tan( 2x)}{x}\cdot{\color{red}\frac{2}{2}}}{\displaystyl e\frac{\sin(3x)}{x}\cdot{\color{red}\frac{3}{3}}})
