Calculus (3rd Edition)

$c=\dfrac{1}{2}$
We have to determine $c$ so that the limit exists: $L=\displaystyle\lim_{x\rightarrow 0}\dfrac{1+cx^2-\sqrt{1+x^2}}{x^4}$ Rationalize the numerator: $L=\displaystyle\lim_{x\rightarrow 0}\dfrac{(1+cx^2-\sqrt{1+x^2})(1+cx^2+\sqrt{1+x^2})}{x^4}$ $=\displaystyle\lim_{x\rightarrow 0}\dfrac{(1+cx^2)^2-(\sqrt{1+x^2})^2}{x^4(1+cx^2+\sqrt{1+x^2})}$ $=\displaystyle\lim_{x\rightarrow 0}\dfrac{1+2cx^2+x^4-1-x^2}{x^4(1+cx^2+\sqrt{1+x^2})}$ $=\displaystyle\lim_{x\rightarrow 0}\dfrac{c^2x^4+(2c-1)x^2}{x^4(1+cx^2+\sqrt{1+x^2})}$ $=\displaystyle\lim_{x\rightarrow 0}\dfrac{x^2[c^2x^2+(2c-1)]}{x^4(1+cx^2+\sqrt{1+x^2})}$ $=\displaystyle\lim_{x\rightarrow 0}\dfrac{c^2x^2+2c-1}{x^2(1+cx^2+\sqrt{1+x^2})}$ As the denominator is 0 for $x=0$, the numerator should also be 0: $c^2(0^2)+2c-1=0$ $2c-1=0$ $c=\dfrac{1}{2}$