Limits
\[\lim_{x \to p} (f(x) + g(x)) = \lim_{x \to p} f(x) + \lim_{x \to p} g(x) \] \[\lim_{x \to p} (f(x) - g(x)) = \lim_{x \to p} f(x) - \lim_{x \to p} g(x) \]When can this be done? For example the limit of \((n^2 - n)\) is not \(\infty - \infty\) but rather just \(\infty\) after simplification.
\[\lim_{x \to p} (f(x) \cdot g(x)) = \lim_{x \to p} f(x) \cdot \lim_{x \to p} g(x) \] \[\lim_{x \to p} \frac{f(x)}{g(x)} = \frac{\lim_{x \to p} f(x)}{\lim_{x \to p} g(x)} \] \[\lim_{x \to p} f(x)^{g(x)} = \lim_{x \to p} f(x)^{\lim_{x \to p} g(x)} \]most often the variable \(x\) approaches the value \(p=\infty\). This then results in the following rules where \(C\) is a constant:
\[\lim_{x \to \infty} C = C \] \[\lim_{x \to \infty} f(x) = C \pm \infty = \pm \infty \] \[\lim_{x \to \infty} f(x) = C \cdot \infty = \begin{cases} \infty & \text{if } C > 0 \\ -\infty & \text{if } C < 0 \end{cases} \]And if it is negative infinity then + and - are swapped.
\[\lim_{x \to \infty} f(x) = \frac{C}{\infty} = 0 \] \[\lim_{x \to \infty} f(x) = \frac{\infty}{C} = \begin{cases} \infty & \text{if } C > 0 \\ -\infty & \text{if } C < 0 \end{cases} \] \[\lim_{x \to \infty} f(x) = \infty^C = \begin{cases} \infty & \text{if } C > 0 \\ 0 & \text{if } C < 0 \end{cases} \] \[\lim_{x \to \infty} f(x) = C^{\infty} = \begin{cases} \infty & \text{if } C > 1 \\ 0 & \text{if } 0 < C < 1 \end{cases} \] \[\lim_{x \to \infty} f(x) = C^{-\infty} = \begin{cases} 0 & \text{if } C > 1 \\ \infty & \text{if } 0 < C < 1 \end{cases} \]L’Hôpital’s Rule
There are some special cases where the limit of a function is not immediately obvious. In these cases we can use L’Hôpital’s Rule to find the limit.
\[\lim_{x \to p} \frac{f(x)}{g(x)} = \lim_{x \to p} \frac{f'(x)}{g'(x)} \]Where \(f'(x)\) and \(g'(x)\) are the derivatives of \(f(x)\) and \(g(x)\) respectively. This rule can be applied multiple times if the limit is still not obvious. Scenarios where this rule is useful are when the limit is of the form \(\frac{0}{0}\) or \(\frac{\pm \infty}{\pm \infty}\).