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 $\color{#3D99F6}{\textbf {coincide the line y=x}}$, thereby giving the first two results in our hand.

Furthermore it's seen that near the origin, the sin curve $\color{#3D99F6}{\textbf { is slightly below the line y=x}}$ & tan curve $\color{#3D99F6}{\textbf { is slightly above the line y=x}}$. 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.. $\Large\text{What's the value of} \color{#D61F06} { \lim_{x\to 0}[\frac{\sin x•\tan x}{x^2}]}$ Can you submit the result decorated with the rigorous proof of yours??

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestis it 1 ?

Log in to reply

Yes indeed.. But share the proof.

Log in to reply

use expansion of sinx and tanx

Log in to reply

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, $\lim_{x \rightarrow 0} [ \frac{\sin(x)\tan(x)}{x^2} ] = 1$

Log in to reply