 # Use Taylor series to evaluate the following limits. lim_{xrightarrow0}frac{sqrt{1+2x}-1-x}{x^2} Maiclubk 2020-11-26 Answered
Use Taylor series to evaluate the following limits.
$\underset{x\to 0}{lim}\frac{\sqrt{1+2x}-1-x}{{x}^{2}}$
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Consider the given limit,
$\underset{x\to 0}{lim}\frac{\sqrt{1+2x}-1-x}{{x}^{2}}$
use Taylor series expansion for the expression $\sqrt{1+2x}$
$\sqrt{1+2x}=\left(1+x-\frac{1}{2}{x}^{2}+\frac{1}{2}{x}^{3}-\frac{5}{8}{x}^{4}...\right)$
Substitute the value into the limit,
$\underset{x\to 0}{lim}\frac{\sqrt{1+2x}-1-x}{{x}^{2}}=\frac{\left(1+x-\frac{1}{2}{x}^{2}+\frac{1}{2}{x}^{3}-\frac{5}{8}{x}^{4}...\right)-1-x}{{x}^{2}}$
$\underset{x\to 0}{lim}\frac{-\frac{1}{2}{x}^{2}+\frac{1}{2}{x}^{3}-\frac{5}{8}{x}^{4}...}{{x}^{2}}$
$\underset{x\to 0}{lim}-\frac{1}{2}+\frac{1}{2}{x}^{-}\frac{5}{8}{x}^{2}...$
$=-\frac{1}{2}$
hence the value of the limit is $-\frac{1}{2}$
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