## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to0}\frac{x\csc2x}{\cos5x}=\frac{1}{2}$$
$$A=\lim_{x\to0}\frac{x\csc2x}{\cos5x}$$ $$A=\lim_{x\to0}\frac{x\times\frac{1}{\sin2x}}{\cos5x}=\lim_{x\to0}\frac{x}{\sin2x\cos5x}$$ $$A=\lim_{x\to0}\frac{x}{\sin2x}\lim_{x\to0}\frac{1}{\cos5x}=\lim_{x\to0}\frac{x}{\sin2x}\times\frac{1}{\cos0}=\lim_{x\to0}\frac{x}{\sin2x}\times\frac{1}{1}$$ $$A=\lim_{x\to0}\frac{x}{\sin2x}$$ $$A=\lim_{x\to0}\Big(\frac{\sin2x}{x}\Big)^{-1}=\Big(\lim_{x\to0}\frac{\sin2x}{x}\Big)^{-1}=X^{-1}$$ Considering $X$: $$X=\lim_{x\to0}\frac{\sin2x}{x}$$ To be able to apply Theorem 7, we need $2x$ in the denominator, instead of $x$. So we would multiply both numerator and denominator by $2$: $$X=\lim_{x\to0}\frac{2\sin 2x}{2\times x}=2\lim_{x\to0}\frac{\sin 2x}{2x}$$ Apply Theorem 7 with $\theta=2x$ here: $$X=2\times1=2$$ Therefore, $$A=X^{-1}=2^{-1}=\frac{1}{2}$$