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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**: $$ \begin{align}\lim_{x\to\infty}\left(\sqrt{x^2+5x+1}-\sqrt{x^2+3x-5}\right)\\=\lim_{x\to\infty}\frac{\left(\sqrt{x^2+5x+1}-\sqrt{x^2+3x-5}\right)\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\=\lim_{x\to\infty}\frac{\left(\cancel{x^2}+5x+1\right)-\left(\cancel{x^2}+3x-5\right)}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\=\lim_{x\to\infty}\frac{2x+6}{\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}}\\=\lim_{x\to\infty}\frac{x(2+\frac6x)}{\sqrt{x^2}\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\=\lim_{x\to\infty}\frac{\cancel{x}(2+\frac6x)}{\cancel{x}\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\=\lim_{x\to\infty}\frac{(2+\frac6x)}{\left(\sqrt{x^2+5x+1}+\sqrt{x^2+3x-5}\right)}\\=\frac{2+0}{\sqrt{1+0+0}+\sqrt{1+0-0}}=1\end{align}$$
- **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**: $$$$