QuestionThe Taylor polynomials we've shown you so far really do closely resemble the functions they come from, close to their centres.
In fact, a degree $n$-Taylor polynomial for $f(x)$ (when $f(x)$ is one of the functions whose Maclaurin polynomials are on the list you should know) at $x=0$ will approximate the function $f(x)$ on the order of $x^{n}$. More precisely, $$\lim _{x \rightarrow a}\left(\frac{f(x)-T_{n}(x)}{x^{n}}\right)=0$$
Notice that the above limit is a $\frac{0}{0}$ form. So the fact that the overall limit is zero means that the numerator $f(x)-T_{n}(x)$ goes to zero faster than the denominator $x^{n}$.
Using this, evaluate the following limits: (a) $\lim _{x \rightarrow 0} \frac{\cos (x)-1+\frac{x^{2}}{2}}{9 x^{4}}=$ $\square$ (b) $\lim _{x \rightarrow 0} \frac{e^{3 x}-1-3 x-\frac{9 x^{2}}{2}}{x^{3}}=$ $\square$ (c) $\lim _{x \rightarrow 0} \frac{\sin (x)-x+\frac{x^{3}}{6}}{2 x^{4}}=$ $\square$ (d) $\lim _{x \rightarrow 0} \frac{e^{x^{5}} \log \left(1+x^{7}\right)+\cos \left(x^{99}\right)-\sin \left(x^{7}\right)-1}{x^{12}}=$ $\square$