> [!summary] Values in determinate forms are special limit values that have a result. > **Determine Forms:** > > | # | Form | Result | | --- | ------------------ | ------------ | | 1 | $\frac{L}{0^-}$ | $\pm \infty$ | | 2 | $\frac{L}{0^+}$ | $\pm \infty$ | | 3 | $\frac{L}{\infty}$ | $0$ | | 4 | $\frac{0}{\infty}$ | $0$ | | 5 | $\frac{\infty}L{}$ | $\infty$ | | 6 | $\infty+\infty$ | $\infty$ | | 7 | $L\cdot \infty$ | $\infty$ | | 8 | $0^\infty$ | $0$ | | 9 | $\infty^\infty$ | $\infty$ | | 10 | $L^\infty$ | $\infty$ | >[!info]+ Read Time **⏱ 2 mins** # Definition Determinate forms are forms of expressions that tell you what the limit is without ambiguity. Therefore can be computed as a limit. ## Derivation > [!warning] Assumptions To show each special determinate form through a limit, assume the following: > - A [[Limits|limit]] is in the form $\displaystyle\lim_{ x \to n } f(x)$ > - A limit is solved using [[Analytical Limits Rules|limit rules]] > - $L$ is a [[Real Numbers|real number]] upset $0$ ($L\in \mathbb{R} \space \textbackslash \left\{ 0 \right\}$) $ \begin{align*} (1) & \displaystyle \lim_{ x \to 0^- } \frac{L}{x} &= \pm \infty \\ (2) & \displaystyle \lim_{ x \to 0^+ }\frac{L}{x} &= \pm \infty \\ (3)& \displaystyle \lim_{ x \to \infty }\frac{L}{x} &= 0 \\ (4)& \displaystyle \lim_{ x \to \infty }\frac{0}{x} &= 0 \\ (5)& \displaystyle \lim_{ x \to \infty }\frac{x}{L} &= \infty \\ (6) & \displaystyle \lim_{ x \to \infty }(x+x) =\displaystyle \lim_{ x \to \infty }(2x) &=\infty \\ (7) & \displaystyle \lim_{ x \to \infty }(L\cdot x) &= \infty\\ (8) & \displaystyle \lim_{ x \to \infty }\left( \frac{1}{x} \right)^x = \displaystyle \lim_{ x \to \infty } (0^x) &= 0 \\ (9) & \displaystyle \lim_{ x \to \infty }x^x &= \infty \\ (10) & \displaystyle \lim_{ x \to \infty }L^x &= \infty \end{align*} $ > [!note] For (1) & (2) if: > $ \begin{array}{c} \text{If $L>0$}, \frac{L}{x} \to \infty \\ \text{If $L<0$}, \frac{L}{x} \to -\infty \end{array} > $