The limit of cosine of three times $x$ minus cosine of four times $x$ divided by $x$ times sine of two times $x$ should be evaluated in this calculus problem as the value of $x$ approaches to zero. Firstly, let’s try to find the limit by direct substitution.
So, substitute $x$ equals to $0$ in the given trigonometric function in rational form.
$=\,\,$ $\dfrac{\cos{3(0)}-\cos{4(0)}}{(0)\sin{2(0)}}$
Now, multiply the constant factors inside each trigonometric function to find their products in the rational function.
$=\,\,$ $\dfrac{\cos{(3 \times 0)}-\cos{(4 \times 0)}}{(0)\sin{(2 \times 0)}}$
$=\,\,$ $\dfrac{\cos{(0)}-\cos{(0)}}{(0)\sin{(0)}}$
$=\,\,$ $\dfrac{\cos{0}-\cos{0}}{0 \times \sin{0}}$
According to trigonometry, the cos of zero radian is one and the sin of zero radian is zero. So, substitute both of them in the rational trigonometric function to find its limit.
$=\,\,$ $\dfrac{1-1}{0 \times 0}$
Now, let’s focus on simplifying this arithmetic expression.
$=\,\,$ $\dfrac{0}{0}$
It is an indeterminate form, which means the limit of $\cos{3x}$ minus $\cos{4x}$ divided $x$ times $\sin{2x}$ cannot be evaluated by direct substitution as the value of $x$ approaches $0$. So, the limit should be evaluated in another method.
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