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2025-11-13 15:01:21 +01:00
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commit ec960e98fa
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.obsidian/workspace.json vendored
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@@ -83,26 +83,12 @@
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18
Gräsvärde (1).md
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@@ -11,11 +11,27 @@
- *Derivator, integraler, asymptot etc definieras med hjälp av gränsvärde.*
- *Om $a$ int är "problempunkt" stoppar vi in $x=a$ i $f(x)$*
- **Def**: *"Problempunkt" t.ex $\lim_{x\to 0}\frac1x$ går inte att direkt lösa på grund av division med $0$*
- **Ex**: $$\begin{align}\lim_{x\to5}f(x)=\lim_{x\to5}\frac1x=\frac15\\\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac1x=0\\\lim_{x\to0}f(x)=\lim_{x\to0}\frac1x\text{ Existerar inte}\end{align}$$
- One sided limits
- ![[gv1.png]]
- **Ex**: $$\begin{align}sgm(x)=\left\{\begin{aligned}1,\;x>0\\0,\;x=0\\-1,\;x<0\end{aligned}\right.\\D_{sgm}=\mathbb{R}\\\lim_{x\to0}sgm(x)\text{ Existerar inte}\\\lim_{x\to0^+}sgm(x)=\lim_{x\to0^+}1=1\\\lim_{x\to0^-}sgm(x)=\lim_{x\to0^-}(-1)=-1\end{align}$$
- **Ex**: $$\begin{align}\lim_{x\to a}f(x)\text{ existerar om}\\\lim_{x\to a+}f(x)\&\lim_{x\to a-}f(x)\\\text{ Existerarf och }\lim_{x\to a+}f(x)=\lim_{x\to a-}f(x)\\\\f(x)=\sqrt{x}, D_f=\left[0,\infty\right)\\\lim_{x\to0+}f(x)=\lim_{x\to0+}\sqrt{x}=0\\\\f(x)=\left\{\begin{aligned}x+1,\;x>0\\0,\;x=0\\2x+1,\ x<0\end{aligned}\right.\\D_f=\mathbb{R}\\\lim_{x\to0+}f(x)=\lim_{x\to0+}x+1\\=0+1=1\\\lim_{x\to0-}f(x)=\lim_{x\to0-}2x+1\\=2\times0+1=0\\\lim_{x\to0}f(x)=1\end{align}$$
- Problem fall
- $\left[\frac00\right]$ form: **Ex**: $$\lim_{x\to1}\frac{x^2-3x+2}{x^2-1},\;\lim_{x\to0}\frac{e^x-1}x,\;\lim_{x\to\infty}\frac{\tan{x}}x$$
- $\left[\frac\infty\infty\right]$ form: **Ex**: $$\lim_{x\to\infty}\frac{x^2-3x+2}{x^2-1},\;\lim_{x\to\infty}\frac{x^3}{2^x}$$
- $\left[0\times\infty\right]$ form: **Ex**: $$\lim_{x\to\infty}x^2\ln\mid{x}\mid$$
- $\left[0^0\right]$ form: **Ex**: $$\lim_{x\to0+}x^x$$
- $\left[\infty^0\right]$ form **Ex**: $$\lim_{x\to\infty}$$
- $\left[\infty^0\right]$ 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**: $$\begin{align}\lim_{x\to1}\frac{x^2-3x+2}{x^2-1}=\frac{0^2-3\times0+2}{0^2-1}=\frac{1+2}{1-1}=\frac{3}{0}\text{ Fins inget gränsvärde}\\\lim_{x\to1}\frac{x^2-3x+2}{x^2-1}\Longleftrightarrow\lim_{x\to1}\frac{(x-1)(x-2)}{(x-1)(x+1)}=\frac{x-2}{x+1}=\frac{1-2}{1+1}=-\frac12\end{align}$$
- **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**: $$$$
- Räkneregler
- *Låt $f$ och $g$ vara funktioner så att $$\lim_{x\to a}f(x)=A,\;\lim_{x\to a}=B,\;\mid{A}\mid<\infty,\;\mid{B}\mid<\infty$$*
- $$\lim_{x\to a}\alpha(f(x)+\beta g(x))=\alpha A+\beta B$$
- $$\lim_{x\to a}f(x)\times g(x)=A\times B$$
- $$\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$$*