# Use Taylor series to evaluate the following limits. lim_{xrightarrow4}frac{ln(x-3)}{x^2-16}

Use Taylor series to evaluate the following limits.
$\underset{x\to 4}{lim}\frac{\mathrm{ln}\left(x-3\right)}{{x}^{2}-16}$
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Cullen
To evaluate the limit using Taylor series,
$\underset{x\to 4}{lim}\frac{\mathrm{ln}\left(x-3\right)}{{x}^{2}-16}$
$\because \mathrm{ln}\left(1+x\right)=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+...$
$\therefore \mathrm{ln}\left(x-3\right)=\mathrm{ln}\left(1+\left(x-4\right)\right)$
$=\left(x-4\right)-\frac{\left(x-4{\right)}^{2}}{2}+\frac{\left(x-4{\right)}^{3}}{3}-\frac{\left(x-4{\right)}^{4}}{4}+...$
$\underset{x\to 4}{lim}\frac{\mathrm{ln}\left(x-4\right)}{{x}^{2}-16}=\frac{\left(x-4\right)-\frac{\left(x-4{\right)}^{2}}{2}+\frac{\left(x-4{\right)}^{3}}{3}-\frac{\left(x-4{\right)}^{4}}{4}+...}{\left(x-4\right)\left(x+4\right)}$
$=\frac{\left(x-4\right)\left[1-\frac{\left(x-4\right)}{2}+\frac{\left(x-4{\right)}^{2}}{3}-\frac{\left(x-4{\right)}^{3}}{4}+...\right]}{\left(x-4\right)\left(x+4\right)}$
$\frac{1-\frac{\left(x-4\right)}{2}+\frac{\left(x-4{\right)}^{2}}{3}-\frac{\left(x-4{\right)}^{3}}{4}+...}{x+4}$
$\frac{1-0+0-0+...}{4+4}$
$=\frac{1}{8}$
$\therefore \underset{x\to 4}{lim}\frac{\mathrm{ln}\left(x-3\right)}{{x}^{2}-16}=\frac{1}{8}$
Jeffrey Jordon