Question

Calculus Help!!
Use a Taylor series to find
lim_{x rightarrow 0}(sqrt{1+x}-1-(x/2))/(4x ^{2})

Answer 1

\[\lim_{x \rightarrow 0}(\sqrt{1+x}-1-(x/2))/(4x ^{2}) \]

Answer 2

\[f(x)=\sqrt{1+x}=(1+x)^{1/2}\]\[f'(x)=\frac{1}{2}(1+x)^{-1/2}\]\[f''(x)=-\frac{1}{4}(1+x)^{-3/2}\]\[f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2}+x^3g(x)\]\[f(x)=1+\frac{x}{2}-\frac{x^2}{8}+x^3g(x)\]\[\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-1-x/2}{4x^2}=\]\[=\lim_{x\rightarrow 0}\frac{1+x/2-x^2/8+x^3g(x)-1-x/2}{4x^2}=\]\[=\lim_{x\rightarrow 0}\frac{-x^2/8+x^3g(x)}{4x^2}=\lim_{x\rightarrow 0}\frac{-1/8+xg(x)}{4}=-\frac{1}{32}\]

Answer 3

Thanks!!