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\sin x<x<\tan x for 0<x<π20< x<\frac{π}{2}

Also it can even be realized by the graphs of sin⁡x\sin x & tan⁡x\tan x that as xx comes closer to origin (orx→0 x\to 0), the graphs of these trigonometric functions coincide the line y=x\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  is slightly below the line y=x\color{#3D99F6}{\textbf { is slightly below the line y=x}} & tan curve  is slightly above the line y=x\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.. What’s the value oflim⁡x→0[sin⁡x•tan⁡xx2]\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??

#Calculus #Graphs #GreatestIntegerFunction(Floor) #Limits #SandwichTheorem

Note by Sanjeet Raria
6 years, 8 months ago

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Comments

is it 1 ?

Parth Lohomi - 6 years, 8 months ago

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Yes indeed.. But share the proof.

Sanjeet Raria - 6 years, 8 months ago

use expansion of sinx and tanx

ojas dhiman - 6 years, 8 months ago

Since this is a even function, it suffice to check

lim⁡x→0+[sin⁡(x)tan⁡(x)x2] \lim_{x \rightarrow 0^{+} } [ \frac{\sin(x)\tan(x)}{x^2} ]

Consider the function f(x)=sin⁡(x)tan⁡(x)−x2f′(x)=sec⁡(x)tan⁡(x)+sin⁡(x)−2x≥2tan⁡(x)−2x≥0 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 f(0^{+}) > f(0)= 0 So sin⁡(x)tan⁡(x)x2≥1 \frac{\sin(x)\tan(x)}{x^2} \ge 1 . Also we see that lim⁡x→0+sin⁡(x)tan⁡(x)x2=1 \lim_{x \rightarrow 0^{+} } \frac{\sin(x)\tan(x)}{x^2} = 1 . So combining these two gives, lim⁡x→0[sin⁡(x)tan⁡(x)x2]=1 \lim_{x \rightarrow 0} [ \frac{\sin(x)\tan(x)}{x^2} ] = 1

Shivang Jindal - 6 years, 7 months ago
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