2.7 Answer

2.7

1.

(a)

$${\lim_{x \to 0}\frac{\sin{2x}}{\sin{3x}} = (\frac{0}{0})}$$ which is indeterminate form. Then by L'Hospital's rule, we have

$$\lim_{x \to 0}\frac{\sin{2x}}{\sin{3x}} = \lim_{x \to 0}\frac{2\cos{2x}}{3\cos{3x}} = \frac{2}{3}$$

(b)

$${\lim_{x \to 0}\frac{\cos{x} - 1}{x} = (\frac{0}{0})}$$ which is indeterminate form. Then by L'Hospital's rule,

$$\lim_{x \to 0}\frac{\cos{x} - 1}{x} = \lim_{x \to 0}\frac{-\sin{x}}{1} = 0$$

(c)

$${\lim_{x \to 0}\frac{\sin^{-1}{x}}{x} = (\frac{0}{0})}$$ which is indeterminate form. Then by L'Hospital's rule,

$$\lim_{x \to 0}\frac{\sin^{-1}{x}}{x} = \lim_{x \to 0}\frac{1}{\sqrt{1 - x^2}} = 1$$

(d)

$$0 \leq \vert x \sin{\frac{1}{x}}\vert = \vert x\vert\vert\sin{\frac{1}{x}}\vert \leq \vert x\vert$$

Then by the squeezing theorem, $${\lim_{x \to 0}x \sin{\frac{1}{x}} = 0}$$

(e)

$${\lim_{x \to 0}(\frac{1}{x^2} - \frac{1}{\sin^{2}{x}}) = \infty - \infty}$$ which is indeterminate form．To use L'Hospital's rule, it must be of the indeterminate form $(\frac{0}{0}),(\frac{\infty}{\infty})$. So, rewrite

$$\lim_{x \to 0}(\frac{1}{x^2} - \frac{1}{\sin^{2}{x}}) = \lim_{x \to 0}\frac{\sin^{2}{x} - x^2}{x^2 \sin^{2}{x}} = (\frac{0}{0})$$

$$= \lim_{x \to 0}\frac{\sin{2x} - 2}{2x \sin^{2}{x} + 2x^2 \sin{2x}} = \lim_{x \to 0}\frac{2\cos{2x} - 2}{2\sin^{2}{x} + 4x \sin{2x} + 2x^2 \cos{2x}}$$

$$= \lim_{x \to 0} \frac{-4\sin{2x}}{6\sin{2x} + 12x \cos{2x} - 4x^2 \sin{2x}}$$

$$= \lim_{x \to 0}\frac{-8\cos{2x}}{24 \cos{2x} - 32 x \sin{2x} + 8x^2 \cos{2x} } = -\frac{1}{3}$$

(f)

$${\lim_{x \to \frac{\pi}{2}}(1 - \sin{x})^{\cos{x}} = (0^{0})}$$ which is indeterminate form. To use L'Hospital's rule, it must be of the indeterminate form $(\frac{0}{0}),(\frac{\infty}{\infty})$．Note that $e^{\log{g(x)}} = g(x)$. Then $${\lim_{x \to \frac{\pi}{2}}(1 - \sin{x})^{\cos{x}} = e^{\lim_{x \to \frac{\pi}{2}}\cos{x}\log{(1 - \sin{x})}} = (0 \cdot \infty)}$$．もう一度変形すると $${e^{\lim_{x \to \frac{\pi}{2}}\cos{x}\log{(1 - \sin{x})}} = e^{\li... ...o \frac{\pi}{2}}\frac{\log{(1 - \sin{x})}}{\frac{1}{\cos{x}}}} = (\frac{0}{0})}$$ よって $${\lim_{x \to \frac{\pi}{2}}\frac{\log{(1 - \sin{x})}}{\frac{1}{\cos{x}}}}$$ にL'Hospitalの定理を用いると

$$e^{\lim_{x \to \frac{\pi}{2}}\frac{\log{(1 - \sin{x})}}{\frac{1}{\cos{x}}}} = e^{\lim_{x \to \frac{\pi}{2}}\frac{\frac{-\cos{x}}{1 - \sin{x}}}{\frac{\sin{x}}{\cos^{2}{x}}}}$$

$$= e^{\lim_{x \to \frac{\pi}{2}}\frac{-\cos^{3}{x}}{\sin{x}(1 - \sin{x})}}$$

$$= e^{\lim_{x \to \frac{\pi}{2}}\frac{3\cos^{2}{x}\sin{x}}{\cos{x} - \sin{2x}}}$$

$$= e^{\lim_{x \to \frac{\pi}{2}}\frac{-6\sin^{2}{x}\cos{x} + 3\cos^{3}{x}}{-\sin{x} - 2\cos{2x}}} = e^{0} = 1$$

(g)

$${\lim_{x \to 0}(\frac{e^{x} - 1}{x})^{1/x} = (1^{\infty})}$$ which is indeterminate form. To use L'Hospital's rule, it must be of the indeterminate form $(\frac{0}{0}),(\frac{\infty}{\infty})$. Note that $e^{\log{g(x)}} = g(x)$. Then $${\lim_{x \to 0}(\frac{e^{x} - 1}{x})^{1/x} = e^{\lim_{x \to 0}\frac{1}{x} \log{(\frac{e^{x} - 1}{x})}} = (\infty \cdot 0)}$$ which is indeterminate form．So, we rewrite $${\lim_{x \to 0}\frac{1}{x} \log{(\frac{e^{x} - 1}{x})} = \lim_{x \to 0}\frac{\log{\frac{e^{x} - 1}{x})}}{x} = (\frac{0}{0})}$$ which is indeterminate form. Then by L'Hospital's rule, we have

$$e^{\lim_{x \to \frac{\pi}{2}}\frac{\log{\frac{e^{x} - 1}{x})}}{x}} = e^{\lim_{x \to \frac{\pi}{2}}\frac{\frac{xe^{x} - e^{x} + 1}{x^2}}{\frac{e^x - 1}{x}}}$$

$$= e^{\lim_{x \to \frac{\pi}{2}}\frac{xe^{x} - e^{x} + 1}{xe^x - x}} = e^{\lim_{x \to \frac{\pi}{2}}\frac{xe^{x}}{xe^x + e^x - x}}$$

$$= e^{\lim_{x \to \frac{\pi}{2}}\frac{xe^{x} + e^{x}}{xe^x + e^x + e^{x}}} = e^{1/2}$$

(h) $e^{-1}$

$${\lim_{x \to 0}(1 - x)^{1/sin{x}} = (1^{\infty})}$$ which is indeterminate form. To use L'Hospital's rule, it must be of the indeterminate form $(\frac{0}{0}),(\frac{\infty}{\infty})$．Note that $e^{\log{g(x)}} = g(x)$. $${\lim_{x \to 0}(1 - x)^{1/sin{x}} = e^{\lim_{x \to 0}\frac{1}{sin{x}} \log{(1 - x)}} = (\infty \cdot 0)}$$ which is indeterminate form．So, rewrite $${\lim_{x \to 0}\frac{1}{\sin{x}} \log{(1 - x)} = \lim_{x \to 0}\frac{\log{1 -x}}{\sin{x}} = (\frac{0}{0})}$$ which is indeterminate form. Thus by L'Hospital's rule, we have

$$e^{\lim_{x \to \frac{\pi}{2}}\lim_{x \to 0}\frac{\log{1 -x}}{\sin{x}}} = e^{\lim_{x \to \frac{\pi}{2}}\frac{\frac{-1}{1-x}}{\cos{x}}} = e^{-1}$$