Replacement Rule

Rule

Let $I$ be an open interval containing the point $x=c$ If $f(x)=g(x)$ for all $I$, except possible at $x=c$, then ${\lim }_{x\to c}f(x)={\lim }_{x\to c}g(x)$

Examples

Simple Example

${\lim }_{x\to 6}\frac{x^2-7x+6}{36-x^2}={\lim }_{x\to 6}\frac{(x-1)(x-6)}{-(6+x)(x-6)}={\lim }_{x\to 6}\frac{(x-1)}{-(6+x)}=-\frac{5}{12}$ This is because $\frac{x-1}{-(6+x)}=\frac{(x-1)(x-6)}{-(6+x)(x-6)}$ except at $x=6$, so we can use $\frac{x-1}{-(6+x)}$ to evaluate $\underset{x\to 6}{\lim }$

If there are square roots involved

\lim_{x \to -3} \frac{\sqrt{x+12} - \sqrt{6-x}}{9-x^2}&& \end{flalign}$$ $$\begin{flalign} =\lim_{x \to -3} \frac{\sqrt{x+12} - \sqrt{6-x}}{(3+x)(3-x)}\ \cdot\ \frac{\sqrt{x+12} + \sqrt{6-x}}{\sqrt{x+12} + \sqrt{6-x}} && \end{flalign}$$ $$\begin{flalign} =\lim_{x \to -3} \frac{(x+12) - (6-x)}{(3+x)(3-x)\left(\sqrt{x+12} + \sqrt{6-x}\right)}=\lim_{x \to -3} \frac{6+2x}{(3+x)(3-x)\left(\sqrt{x+12} + \sqrt{6-x}\right)}=&& \end{flalign}$$ $$\begin{flalign} =\lim_{x \to -3} \frac{2}{(3-x)\left(\sqrt{x+12} + \sqrt{6-x}\right)} =\frac{2}{(3+3)\left(\sqrt{12-3} + \sqrt{6+3}\right)} =\frac{1}{18}&& \end{flalign}$$