数学の基礎と公式-15
数学の基礎と公式
amazon kindle版を出版しました。
15. 極限-1
$\displaystyle{ }$
(1) $\displaystyle{\lim_{x\rightarrow 0}{\dfrac{\sin x}{x}=1}}$
(2) $\displaystyle{\lim_{x \rightarrow \infty}{\dfrac{3^x+2^x}{4^x}}=\lim_{x \rightarrow 0} \left \lbrace \left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n \right\rbrace =0}$
(3) $\displaystyle{\lim_{x \rightarrow \infty}{\dfrac{2^{2x-1}}{3^x-1}}=\lim_{x \rightarrow \infty}\left \lbrace\dfrac{\dfrac{1}{2 }\left(\dfrac{4}{ 3}\right)^x}{1-\dfrac{1}{3^x}}\right\rbrace =+\infty}$
(4) $\displaystyle{\lim_{x \rightarrow \infty} (2^x-4^x)=\lim_{x \rightarrow \infty}4^x \left( \dfrac{2^x}{4^x}-1 \right)=-\infty }$
(5) $\displaystyle{\lim_{x \rightarrow \infty} \dfrac{5x^2+3x+1}{2x^2+4x+5} =\lim_{x \rightarrow \infty} \dfrac{5+\dfrac{3}{x}+\dfrac{1}{x^2}}{2+\dfrac{4}{x}+\dfrac{5}{x^2}}=\dfrac{5}{2}}$
(6) $\displaystyle{\lim_{x \rightarrow \infty}\dfrac{\sqrt{x^2-5x}-3x}{x} =\lim_{x \rightarrow \infty}\dfrac{\sqrt{1-\dfrac{5}{x}}-3}{1}=-2 }$
(7) $\displaystyle{\lim_{x \rightarrow \infty}\dfrac{1}{x} \sin \dfrac{x %pi}{2}=0 }$
(8) $\displaystyle{\lim_{x \rightarrow \infty} \{ \log(x+1)-\log x \}}$
$\displaystyle{=\lim_{x \rightarrow \infty} \{ \log \dfrac{x+1}{x} \}=\lim_{x \rightarrow \infty} \{ \log \dfrac{1+\dfrac{1}{x}}{1} \}=0 }$
(9) $\displaystyle{\lim_{x \rightarrow \infty} \{ \log(4x+1)-\log(2x-1) \} }$
$\displaystyle{= \lim_{x \rightarrow \infty} \left \lbrace \log \dfrac{4x+1}{2x-1} \right \rbrace= \lim_{x \rightarrow \infty} \left \lbrace \log \dfrac{4+\dfrac{1}{x}}{2-\dfrac{1}{x}} \right \rbrace=2 }$
(10) $\displaystyle{ \lim_{x \rightarrow 0} \left( \dfrac{\sqrt{1+2x+3x^2}-\sqrt{1-3x}}{3x} \right) }$
$\displaystyle{=\lim_{x \rightarrow 0} \left( \dfrac{{1+2x+3x^2}-{1-3x}}{3x(\sqrt{1+2x+3x^2}+\sqrt{1-3x})} \right) }$
$\displaystyle{=\lim_{x \rightarrow 0} \left( \dfrac{5x+3x^2}{3x(\sqrt{1+2x+3x^2}+\sqrt{1-3x})} \right) =\dfrac{5}{6} }$
(11) $\displaystyle{ \lim_{x \rightarrow \infty} (\sqrt{x^2+x+1}-\sqrt{x^2+1}) }$
$\displaystyle{= \lim_{x \rightarrow \infty} \left(\dfrac{x^2+x+1-(x^2+1)}{\sqrt{x^2+x+1}+\sqrt{x^2+1}} \right) }$
$\displaystyle{= \lim_{x \rightarrow \infty} \left(\dfrac{x}{\sqrt{x^2+x+1}+\sqrt{x^2+1}} \right) }$
$\displaystyle{= \lim_{x \rightarrow \infty} \left(\dfrac{1}{\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}+\sqrt{1+\dfrac{1}{x^2}}} \right)=\dfrac{1}{2} }$
(12)$\displaystyle{ \lim_{x \rightarrow \pm \infty} \left( 1+\dfrac{1}{x} \right)^x }$
$=\displaystyle{ \lim_{x \rightarrow \pm \infty} \exp \log \left[ \left( 1+\dfrac{1}{x} \right)^x \right] }$
$=\displaystyle{\exp \lim_{x \rightarrow \pm \infty} \log \left[ \left( 1+\dfrac{1}{x} \right)^x \right] }$
ここで、$x=\dfrac{1}{X}$とおくと、
$=\displaystyle{\exp \lim_{X \rightarrow \pm 0} \log \left[\left( 1+X \right)^{\frac{1}{X}} \right] }$
$=\displaystyle{\exp \lim_{X \rightarrow \pm 0} \dfrac{\log \left( 1+X \right)}{X} }$
ロピタルの定理から
$=\displaystyle{\exp \lim_{X \rightarrow \pm 0} \dfrac{1}{1+X} }=\exp(1)=e$
(13) $\displaystyle{ \lim_{x \rightarrow \infty} \left( 1+\dfrac{1}{2x} \right)^x }=\displaystyle{ \lim_{x \rightarrow \infty} \left( 1+\dfrac{1}{2x} \right)^{2x \times \frac{1}{2}} }$
(12)式より
$=e^\frac{1}{2}$
(14) $\displaystyle{ \lim_{x \rightarrow \infty} \left( 1-\dfrac{2}{x} \right)^x }=\displaystyle{ \lim_{x \rightarrow \infty} \left( 1+\dfrac{1}{\frac{x}{-2}} \right)^{\frac{x}{-2}\times(-2)} }$
(12)式より
$=e^{-2}$
(15) $\displaystyle{ \lim_{x \rightarrow \infty} \left( \dfrac{x}{x+1} \right)^x }$
$=\displaystyle{ \lim_{x \rightarrow \infty} \left( \dfrac{x+1-1}{x+1} \right)^x }$
$=\displaystyle{ \lim_{x \rightarrow \infty} \left( 1-\dfrac{1}{x+1} \right)^x }$
$=\displaystyle{ \lim_{x \rightarrow \infty} \left( 1+\dfrac{1}{-(x+1)} \right)^{\{-(x+1)\times (-1)\}} \times \left( 1+\dfrac{1}{-(x+1)} \right) }$
(12)式より
$=e^{-1}$
(16) $\displaystyle{ \lim_{x \rightarrow 0} \left( 1+zx \right)^\frac{1}{x} }$
$t=zx$とおくと、
$=\displaystyle{ \lim_{t \rightarrow 0} \left( 1+t \right)^{\frac{1}{t} \times z} }$
(12)式より
$=e^{z}$
(17) $\displaystyle{ \lim_{x \rightarrow 0} \left( 1-x^2 \right)^\frac{1}{x} }$
$\displaystyle{ =\lim_{x \rightarrow 0} \{( 1-x)(1+x)\}^\frac{1}{x}=\lim_{x \rightarrow 0} ( 1-x)^{\frac{1}{-x}\times(-1) }(1+x)^\frac{1}{x} }$
(12)式より
$=e^{-1}\cdot e=1$
maximaでの記述
(1)$\displaystyle{\lim_{x\rightarrow 0}{\dfrac{\sin x}{x}=1}}$
limit(sin(x)/x, x, 0);
1
(2)$\displaystyle{\lim_{x\rightarrow +0}{x^{x}=1}}$
limit(x^x, x, 0,plus);
1
$x\rightarrow +0$のときplus
$x\rightarrow -0$のときminus
