Question #2880c

Dec 14, 2017

Multiply the top and bottom by the fraction $\frac{60 x}{60 x}$ before rearranging and using the given limit to obtain an answer of $\frac{20}{3}$.

Explanation:

Let $f \left(x\right) = \frac{\sin \left(5 x\right) \cot \left(3 x\right)}{x \cot \left(4 x\right)}$. Then

$f \left(x\right) = \frac{60 x}{60 x} \cdot f \left(x\right) = \frac{60 x \sin \left(5 x\right) \cot \left(3 x\right)}{60 {x}^{2} \cot \left(4 x\right)}$

$= \frac{60 x \sin \left(5 x\right) \cos \left(3 x\right) \sin \left(4 x\right)}{60 {x}^{2} \sin \left(3 x\right) \cos \left(4 x\right)}$

$= \frac{3 x}{\sin \left(3 x\right)} \cdot \sin \frac{5 x}{5 x} \cdot \sin \frac{4 x}{4 x} \cdot \frac{20 \cos \left(3 x\right)}{3 \cos \left(4 x\right)}$

Since ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$, it follows that ${\lim}_{x \to 0} \frac{3 x}{\sin \left(3 x\right)} = 1$, ${\lim}_{x \to 0} \sin \frac{5 x}{5 x} = 1$, and ${\lim}_{x \to 0} \sin \frac{4 x}{4 x} = 1$.

Moreover, we can say that ${\lim}_{x \to 0} \frac{20 \cos \left(3 x\right)}{3 \cos \left(4 x\right)} = \frac{20 \cdot 1}{3 \cdot 1} = \frac{20}{3}$ by substitution into the function since $\frac{20 \cos \left(3 x\right)}{3 \cos \left(4 x\right)}$ is continuous at $x = 0$.

Finally, applying the rule for the limit of a product of functions (whose individual limits exist), we obtain

${\lim}_{x \to 0} f \left(x\right) = 1 \cdot 1 \cdot 1 \cdot \frac{20}{3} = \frac{20}{3}$.