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What is the limit, using the Taylor series: lim as x approaches zero, ((sqrt (1+x))-1-(x/2)) / 4x^2
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do you know the taylor series for \(\sqrt{1+x}\) ?
actually you only need the first three terms
\[\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}\] subtracting leaves \(-\frac{x^2}{8}\)
then divide by \(4x^2\) and that will be your limit
alright, I have not used this site before. I am looking it over. Thanks a lot I have many challenging calc II questions
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you got this? when you divide you get a constant all other higher powers will contain \(x^k\) so when you let \(x=0\) they are gone and you are left only with the constant
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