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About LinkBacksokay 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
We should use the fact that $\displaystyle \lim_{u\to0}\frac{\tan u}{u}=1$ and $\displaystyle \lim_{u\to0}\frac{\sin u}{u}=1$Originally Posted by veezey11
Thus, $\displaystyle \lim_{x\to 0}\frac{\tan(2x)}{\sin(3x)\cos(4x)}=\lim_{x\to0}\f rac{\tan(2x)}{\sin(3x)}\cdot\underbrace{\lim_{x\to 0}\frac{1}{\cos(4x)}}_1=\lim_{x\to0}\frac{\display style\frac{\tan(2x)}{x}}{\displaystyle\frac{\sin(3 x)}{x}}=\lim_{x\to0}\frac{\displaystyle\frac{\tan( 2x)}{x}\cdot{\color{red}\frac{2}{2}}}{\displaystyl e\frac{\sin(3x)}{x}\cdot{\color{red}\frac{3}{3}}}$ $\displaystyle =\lim_{x\to0}\frac{2\displaystyle\frac{\tan(2x)}{2 x}}{3\displaystyle\frac{\sin(3x)}{3x}}=\frac{2}{3} \frac{\displaystyle\lim_{x\to0}\frac{\tan(2x)}{2x} }{\displaystyle\lim_{x\to0}\frac{\sin(3x)}{3x}}=\f rac{2}{3}\cdot\frac{1}{1}=\color{red}\boxed{\frac{ 2}{3}}$
Does this make sense?
--Chris
The easier way is with L'Hospital's rule, namely
$\displaystyle \lim_{x \to a} \frac{g(x)}{h(x)} = \lim_{x \to a} \frac{g'(x)}{h'(x)}$.
So $\displaystyle \lim_{x \to 0} \frac{\tan{2x}}{\sin{3x}\cos{4x}} = \lim_{x \to 0} \frac{\frac{d}{dx}(\tan{2x})}{\frac{d}{dx}(\sin{3x }\cos{4x})}$
$\displaystyle = \lim_{x \to 0} \frac{\frac{2}{\cos^2{2x}}}{3\cos{3x}\cos{4x} - 4\sin{3x}\sin{4x}}$
$\displaystyle = \lim_{x \to 0} \frac{2}{\cos^2 {2x}[3\cos{3x}\cos{4x} - 4\sin{3x}\sin{4x}]}$
Can you go from here?
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