Limits And The Greatest Integer Function

Using the following limit $\displaystyle \lim_{x\rightarrow 0} \frac{\sin x }{x}=1$ We have to modify the expression $\displaystyle \lim_{x\rightarrow 0} \frac{\sin x \cdot \tan x}{x^2} =\lim_{x\rightarrow 0} \frac{\sin x \cdot \frac{\sin x}{\cos x}}{x^2} =\lim_{x\rightarrow 0} \frac{(\sin x)^2 }{x^2\cdot \cos x} =\\ \displaystyle \lim_{x\rightarrow 0} \frac{\sin x}{x}\cdot \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1$

actually $sinx=x-x^3/3!+....$ and $tanx=x+x^3/3+....$ therefore $sinx \times tanx=x^2+x^4/3!+....$ therefore $sinx \times tanx$ is greater 1 when x tends to 0

Observing the graph, we get the answer $1$ . $_\square$

the only confusion in this problem is whether the equation sin(x)/x * tan(x)/x is slightly greater or less than one .... so we compare at what rate the both equations(sinx/x , tanx/x) tend to 1 . sinx/x applying l'hopital rule we get cosx ,which means that the equation is tending to one at the rate cosx , similarly for tanx/x we get sec^2(x) . so in the product of both the equations .... both tending to 1 but one along cosx , and other along sec^2(x), gives us sec(x) which is always greater than one(which is bigger than rate of x) ,hence the product is slightly greater than 1 .

Use of expansion of sinx and tanx works here as we need to apply floor

Use Maclaurin Expansion for $\sin x$ and $\tan x.$ $\sin x = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!} - \cdots$ $\tan x = x+\dfrac{x^3}{3}+\dfrac{2x^5}{15} + \cdots$ So $\dfrac{\sin x \tan x}{x^2} \approx 1+\dfrac{x^2}{6} \rightarrow 1^{+}$ . So the floor of this limit is 1.

My method is crude but very simple and visual. At x -> 0, sinX,tanX ~~x(similar to). If we imagine an infinitely narrow observation, sinX & tanX both looks like what we see in y=x graph. So using this method, lim x->0, sinXtanX/x^2 = x^2/x^2 = 1.

Btw tanX comes to have similar property with sinX because cosX finds the look of y=1 constant graph.

By some calculus, $\frac{\sin x\tan x}{x^2}>1\forall x\in(-\frac{\pi}{6},\frac{\pi}{6})$ \ { $0$ }, then use squeeze lemma to get limit =1

Form 0/0. Apply L'Hopital's rule. Still 0/0 apply it again. Get 2/2 = 1.

With a bit of moderately heavy lifting with the product rule....

After first application it is Limit x->0 (sin(x) + sec(x)tan(x))/2x.

After second application it is Limit x->0 (cos(x) + sec^3(x) + sec(x)tan*2(x)}/2.

This evaluates straightforwardly to (1 + 1 + 0) /2 = 1