$\lim_{x \to 0}\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\rfloor=?$

Using the Taylor series $$ \begin{aligned} \tan x&=x+\frac13x^3+\frac2{15}x^5+\cdots,\\ \sin x&=x-\frac16x^3+\frac1{120}x^5+\cdots \end{aligned} $$ and the binomial formula, we arrive at $$ \tan^{98}x=x^{98}+\frac{98}3x^{100}+[\frac19\cdot\binom{98}2+\frac{98\cdot2}{15}]x^{102}+\cdots $$ and $$ \sin^{98}x=x^{98}-\frac{98}6x^{100}+[\frac1{36}\cdot\binom{98}2+\frac{98}{120}]x^{102}+\cdots. $$ Therefore $$ \tan^{98}x-\sin^{98}x=49x^{100}+\frac{1225}3x^{102}+\cdots, $$ and $$ \frac{\tan^{98}x-\sin^{98}x}{x^{100}}=49+\frac{1225}3x^2+\cdots. $$ When $x$ is close enough to zero, that quadratic term dominates the (possibly negative) cut-off terms. Therefore, when $|x|$ is small enough we have $$ 49\le \frac{\tan^{98}x-\sin^{98}x}{x^{100}}<50. $$ The answer is thus $49$.

No, your last transformation (taking the floor in and out of the limit) is not justified.

As a counterexample, consider

$$\lim_{n\to\infty}\left(49-\frac1n\right)=49$$

versus

$$\lim_{n\to\infty}\left\lfloor49-\frac1n\right\rfloor=48.$$