## University Calculus: Early Transcendentals (3rd Edition)

$L(x)=x$ and $Q(x)=x$
We have $f(x) = \sin x$ and $f'(x)= \cos x ; \\f''(x)= - \sin x$ $\implies f(0)=0; f'(0)=1; f''(0)=0$ Therefore, $L(x)=f(0)+xf'(0)=x$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2}{2!}f''(0)=x$ So, $L(x)=x$ and $Q(x)=x$