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# Oh boy! Can't tell you the limit

I guess everyone seems to be knowing the old results,

$\Large\bullet \lim_{x\to 0}\frac{\sin x}{x}=1$ $\Large\bullet \lim_{x\to 0}\frac{\tan x}{x}=1$ $\Large\bullet \lim_{x\to 0}[\frac{\sin x}{x}]=0$ $\Large\bullet \lim_{x\to 0}[\frac{\tan x}{x}]=1$ $$(\textbf {[•] is greatest integer function})$$

Yeah you know them, great. One can easily get them straight from the inequality

$$\sin x<x<\tan x$$ for $$0< x<\frac{π}{2}$$

Also it can even be realized by the graphs of $$\sin x$$ & $$\tan x$$ that as $$x$$ comes closer to origin (or$$x\to 0$$), the graphs of these trigonometric functions $$\color{blue}{\textbf {coincide the line y=x}}$$, thereby giving the first two results in our hand.

Furthermore it's seen that near the origin, the sin curve $$\color{blue}{\textbf { is slightly below the line y=x}}$$ & tan curve $$\color{blue}{\textbf { is slightly above the line y=x}}$$. Did you smell the proof of the last two of the old results?

Anyways, everything's real easy till we reach to the puzzle of ours.. $\Large\text{What's the value of} \color{red} { \lim_{x\to 0}[\frac{\sin x•\tan x}{x^2}]}$ Can you submit the result decorated with the rigorous proof of yours??

Note by Sanjeet Raria
3 years, 2 months ago

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## Comments

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Since this is a even function, it suffice to check

$\lim_{x \rightarrow 0^{+} } [ \frac{\sin(x)\tan(x)}{x^2} ]$

Consider the function $f(x) = \sin(x)\tan(x)-x^2 \\ f'(x) = \sec(x)\tan(x) + \sin(x) - 2x \ge 2\tan(x)-2x \ge 0$(BY AM-GM inequality. So this function is increasing. This means that, $$f(0^{+}) > f(0)= 0$$ So $\frac{\sin(x)\tan(x)}{x^2} \ge 1$ . Also we see that $\lim_{x \rightarrow 0^{+} } \frac{\sin(x)\tan(x)}{x^2} = 1$ . So combining these two gives, $\lim_{x \rightarrow 0} [ \frac{\sin(x)\tan(x)}{x^2} ] = 1$

- 3 years, 1 month ago

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use expansion of sinx and tanx

- 3 years, 2 months ago

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is it 1 ?

- 3 years, 2 months ago

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Yes indeed.. But share the proof.

- 3 years, 2 months ago

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