can someone prove that 1=sin(x)/x

No one can prove something false

>>9152192
Yes.
[math]\frac{\sin\pi}{\pi} = 0 = 1\;\;\;Q.E.D.[/math]

I don't think thats true anon.

>>9152192
1. Your statement is false in most cases.
2. If you mean lim_(x->0) sin(x)/x = 1 then use l'Hospital

>>9152227
yeah, forgot to include the limit there, but lhospital wouldnt work

>>9152231
Why not? It's the "0/0" case, l'Hospital does work in that case,

>>9152192
shit taste in waifus

use a triangle and sin(x) =opposite/adjacent, then becomes mid school algebra.

File: 1501545826966.gif (2MB, 260x192px) Image search: [Google]

2MB, 260x192px

>>9152231
>lhospital

>>9152192
>Use l'Hopital's rule
[eqn] \lim _{ x \to 0 } \cos ( x ) = 1 [/eqn]

>Recognise that it's a removable singularity
Expand [math] \sin ( x ) [/math] as its Taylor series [eqn] \frac { 1 } { x } \left ( x - \frac { x^2 } { 2! } + \cdots \right ) = 1 - \frac { x } { 2! } [/eqn] then the limit becomes trivial.

>Use the Squeeze theorem
Note that [eqn] \cos ( x ) \leq \frac { \sin ( x ) } { x } \leq 1 [/eqn] then [eqn] \lim _{ x \to 0 } \cos ( x ) \leq \lim _{ x \to 0 } \frac { \sin ( x ) } { x } \leq 1 \\ \implies 1 \leq \lim _{ x \to 0 } \frac { \sin ( x ) } { x } \leq 1 \\ \implies \lim _{ x \to 0 } \frac { \sin (x) } { x } = 1 [/eqn]

>>9152227
it's l'Ambulance you mongoloid