vault backup: 2025-11-13 15:59:21

Gräsvärde (1).md

@@ -33,8 +33,7 @@
 =\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}
-$$
+\end{align}$$
 	- **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**: $$$$