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??

## Comments

Sort by:

TopNewestSince 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, \[ \lim_{x \rightarrow 0} [ \frac{\sin(x)\tan(x)}{x^2} ] = 1 \] – Shivang Jindal · 2 years, 11 months ago

Log in to reply

use expansion of sinx and tanx – Ojas Dhiman · 2 years, 11 months ago

Log in to reply

is it 1 ? – Parth Lohomi · 2 years, 11 months ago

Log in to reply

– Sanjeet Raria · 2 years, 11 months ago

Yes indeed.. But share the proof.Log in to reply