lim sin(x)/x

[delicate marlin]

when u try to prove this limit, u do sin(x) <= x <= tan(x). Why is this true?

[hoary basin]

because x is in the middle of those 2, you can use the squeeze/sandwich theorem

sin divided by all of those, you get:

1 <= sin(x)/x <= cos(x)

this is valid when sin is positive

hm, maybe not quite

valid when sin and tan are positive

and x

[spice sleet]

Really bad picture, but that's the geometric reasoning

big triangle is red, sector is orange, small triangle is blue

[twin pumice]

Here's a similar diagram stolen from https://math.stackexchange.com/a/75151 (a link which contains many answers to your question)

when x is in radians, the area of the inner triangle is 1/2 sin x, the area of the wedge (colored cyan) is 1/2 x, and the area of the outer triangle is 1/2 tan x

hence 1/2 sin x ≤ 1/2 x ≤ 1/2 tan x

[spice sleet]

oh yeah it's the lengths not the areas

[delicate marlin]

well, rlly sorry for my late response

yeah, i know thats the squeeze theorm

But when x (angle) is small, what guarantees me that x (arc length now) is in between of those?

A draw isnt usefull here

Maybe the tan is shorter than x(arc length)

[twin pumice]

what are the areas of the red and green triangles?

in terms of x

[delicate marlin]

red is cos * sin /2 and green is tan / 2

[twin pumice]

right, sorry, wrong red triangle

how about now?

[delicate marlin]

sin / 2

but my question is: what guarantees me that the green isnt as big as the blue one. After all, the green is a curve

[twin pumice]

and what about the blue slice?

[delicate marlin]

it is clearly longer than the red one, since both have the same height but green is a curve

but why shorter than the blue one?

[twin pumice]

I mean in this image

what's the area of the circular slice?

[delicate marlin]

u could "flat" the green curve and maybe it will be same or more large than the blue one

[delicate marlin]

x/2pi iirc

[twin pumice]

not quite

x is in radians

so we have x/2pi part of the whole circle

[delicate marlin]

srry, x*r^2/pi

if r = 1 then x/pi

no?

and x angle in radians

[twin pumice]

the area is π r²

[delicate marlin]

so in this case it is x/pi

[twin pumice]

the pis cancle out

[delicate marlin]

thats for the whole circle, aka 2*pi

[twin pumice]

yes

and the fraction of the circle we have is (\frac{x}{2 \pi})

[shell saffron]

Invariance

[delicate marlin]

yeah

[twin pumice]

so (\frac{x}{2 \pi} \cdot \pi r^2) is the area of the slice

[shell saffron]

Invariance

[delicate marlin]

oh x/2

[twin pumice]

yes!

[delicate marlin]

so the area of the red triangle is smaller than the circular sector

[twin pumice]

clearly, the area of the green triangle is greater than the area of the slice, and the area of the slice is greater than the area of the red triangle

we've calculated those areas

write out the inequality in terms of x

[delicate marlin]

so we are comparing areas in the end mmm

would be sin/2 < x/2 < tan/2 -> sin < x < tan

i see

[twin pumice]

yep!

[twin pumice]

there are several other proofs at this link

if you're interested

(including that one)

[delicate marlin]

i see, i didnt know we were comparing areas

thanks 😄

[twin pumice]

sorry if I didn't make that clear

[twin pumice]

happy to help!

[delicate marlin]

ye ye

it was clear lol

i though we were comparing the length

not the areas

but yes, if 3 "triangles" have the same base with different areas...

Only the height must be different