$\lim_{x \to 0}\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\rfloor=?$ fine the limit :
$$\lim_{x \to 0}\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\rfloor=?$$
We denote the floor funtion by $\lfloor x\rfloor$.
My try:
\begin{align}
\lim_{x \to 0}\frac{\tan^{n}x - \sin^{n} x}{x^{n + 2}} &= \lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}\cdot \sum_{i = 0}^{n - 1}\frac{\tan^{n - 1 - i}x}{x^{n - 1 - i}}\cdot\frac{\sin ^{i}x}{x^{i}}\\
&= \frac{1}{2}\cdot\sum_{i = 0}^{n - 1}1\\
&= \frac{n}{2}
\end{align}
So:
\begin{align}
\lim_{x \to 0}\frac{\tan^{98}x - \sin^{98} x}{x^{100}}&=49\\
\lim_{x \to 0}\left\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\right\rfloor&=49 
\end{align}
is this correct?
 A: 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 $49$.
A: 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
$$\lim_{n\to\infty}\left\lfloor49-\frac1n\right\rfloor=48.$$
