Answer
$c=\dfrac{1}{2}$
Work Step by Step
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:
$c^2(0^2)+2c-1=0$
$2c-1=0$
$c=\dfrac{1}{2}$
