# Use Taylor series to evaluate the following limits. Express the result in terms of the nonzero real parameter(s). lim_{xrightarrow0}frac{e^{ax}-1}{x}

Use Taylor series to evaluate the following limits. Express the result in terms of the nonzero real parameter(s).
$\underset{x\to 0}{lim}\frac{{e}^{ax}-1}{x}$
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joshyoung05M
The given limit $\underset{x\to 0}{lim}\frac{{e}^{ax}-1}{x}$ can be evaluated as,
The Taylor series expansion of ${e}^{ax}$ is,
${e}^{ax}=1+\frac{ax}{1!}+\frac{\left(ax{\right)}^{2}}{2!}+\frac{\left(ax{\right)}^{3}}{3}+...$
Therefore,
$=\underset{x\to 0}{lim}\left(\frac{{e}^{ax}-1}{x}\right)=\underset{x\to 0}{lim}\left(\frac{\left(1+\frac{ax}{1!}+\frac{\left(ax{\right)}^{2}}{2!}+\frac{\left(ax{\right)}^{3}}{3}+...\right)-1}{x}\right)$
$=\underset{x\to 0}{lim}\left(a+\frac{{a}^{2}x}{2!}+\frac{{a}^{3}{x}^{2}}{3!}+...\right)$
$=a$
Hence, $\underset{x\to 0}{lim}\frac{{e}^{ax}-1}{x}=a$
Jeffrey Jordon