## Thomas' Calculus 13th Edition

$L(x)=1$ and $Q(x)=1+\dfrac{x^2}{2!}$
Differentiate the given function $f(x)$ as follows: $f'(x)= \sinh x ; \\f''(x)= \cosh x$ Now $f(0)=1; \\ f'(0)=1; \\ f''(0)=1$ Therefore, $L(x)=f(0)+xf'(0)=1$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2 f''(0)}{2!}=1+\dfrac{x^2}{2!}$