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Gräsvärde (1).md
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- $\left[\infty^0\right]$ form **Ex**: $$\lim_{x\to\infty}x^{1/x}$$
- $\left[1^\infty\right]$ form: **Ex**: $$\lim_{x\to0}(1+x)^{1/x}$$
- $\left[\infty-\infty\right]$ form: $$\lim_{x\to\infty}\left(\sqrt{x^2+5x+1}-\sqrt{x^2+3x-5}\right)$$
- **Ex**: $$
\begin{align}
\lim_{x\to\infty}\left(\sqrt{x^2+5x+1}-\sqrt{x^2+3x-5}\right)\\
=\lim_{x\to\infty}\frac{\left(\sqrt{x^2+5x+1}-\sqrt{x^2+3x-5}\right)\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\
=\lim_{x\to\infty}\frac{\left(\cancel{x^2}+5x+1\right)-\left(\cancel{x^2}+3x-5\right)}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\
=\lim_{x\to\infty}\frac{2x+6}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\
=\lim_{x\to\infty}\frac{x(2+\frac6x)}{\sqrt{x^2}\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\
=\lim_{x\to\infty}\frac{\cancel{x}(2+\frac6x)}{\cancel{x}\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\
=\lim_{x\to\infty}\frac{(2+\frac6x)}{\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\
=\frac{2+0}{\sqrt{1+0+0}+\sqrt{1+0-0}}=1
\end{align}
$$
- **Ex**: $$\begin{align}\lim_{x\to1}\frac{x^2-3x+2}{x^2-1}=\frac{0^2-3\times0+2}{0^2-1}=\frac{1+2}{1-1}=\frac{3}{0}\text{ Fins inget gränsvärde}\\\lim_{x\to1}\frac{x^2-3x+2}{x^2-1}\Longleftrightarrow\lim_{x\to1}\frac{(x-1)(x-2)}{(x-1)(x+1)}=\frac{x-2}{x+1}=\frac{1-2}{1+1}=-\frac12\end{align}$$
- **Ex**: $$\lim_{x\to\infty}\frac{x^2-3x+2}{x^2-1}=\lim_{x\to\infty}\frac{1-\frac3x+\frac2{x^2}}{1-\frac1{x^2}}=\frac{1-0+0}{1-0}=1$$
- **Ex**: $$$$
@@ -34,4 +46,8 @@
- $$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{A}{B}\text{ om }B\neq0$$
- **Theorem**: *Instängningsregel $$\left.\begin{aligned}f(x)\leq g(x)\leq h(x),\;\forall x\\\lim_{x\to a}f(x)=L=\lim_{x\to a}h(x)\end{aligned}\right\}\Rightarrow\lim_{x\to a}g(x)=L$$*
- **Theorem**: $$f(X)\leq g(x),\;\forall x\Rightarrow\;\lim_{x\to a}f(x)\leq\lim_{x\to a}g(x)$$
- **Theorem**: *Sammansättningsregel $$\left.\begin{aligned}\lim_{x\to a}f(x)=b\\\lim_{x\to b}g(x)=L\end{aligned}\right\}\Leftarrow$$*
- **Theorem**: *Sammansättningsregel $$\left.\begin{aligned}\lim_{x\to a}f(x)=b\\\lim_{x\to b}g(x)=L\end{aligned}\right\}\Rightarrow\lim_{x\to a}g\circ f(x)=L$$*
- **Variabelbyte**: $$\lim_{x\to a}g\circ f(x)=\lim_{t\to b}g(x)\text{ där }t=f(x)\longrightarrow b\text{ då }x\longrightarrow a$$
- **Ex**: $$
\begin{align}\lim_{x\to0}x^2\sin\frac1x\\-1\leq\sin\frac1x\leq1,\; x\neq0\\\Rightarrow-x^2\leq x^2\sin\frac1x\leq x^2\\\lim_{x\to0}-x^2=0=\lim_{x\to0}x^2\\\text{Enlight instängningsregel, }\\\lim_{x\to0}x^2\sin\frac1x=0\\\end{align}
$$