Derivative of sinx from first principle

Asked Nov 01 '22 at 18:12

Active Nov 01 '22 at 18:30

Viewed 46 times

I am just confused over this step in the derivative of sin x

$$ \lim_{x \to 0}{\sin\left(x\right) \over x} = 1 $$

When we have the limit, my textbook uses the small angle approximations, to say $\sin\left(x\right) \approx x$ for small $x$ in radians.
However, how is it that we can simply replace $\sin\left(x\right)$ with $x$, and keep the limit still there, surely we must have taken the limit to reach the conclusion that $\sin\left(x\right) \approx x$, and thus we cannot keep the limit there, as we have already applied $\mbox{it}~?$.

The limit is kept there to reach the conclusion it equals $1$.

edited Nov 01 '22 at 18:30

Felix Marin

asked Nov 01 '22 at 18:12

Nav Bhatthal

1

Seems a little circular. What book are you using? – Andrew Chin Nov 01 '22 at 18:15
Edexcel A Level Mathematics Pure year 2 – Nav Bhatthal Nov 01 '22 at 18:16
The standard approach is to use the squeeze theorem and an inequality using areas of part of a circle and some related triangle areas. Here's a video going over it https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-8/v/sinx-over-x-as-x-approaches-0 – N8tron Nov 01 '22 at 18:17
No it is not like this that it works : it should be considered as a consequence of this classical "squeezing" $\forall x \in (0,\pi/2), \sin x <x<\ \tan x \implies (\sin x/x) <1<(\sin x/x) \times (1/\cos x) \implies \cos x < (\sin x/x)<1$ where the second double inequality comes from division by $x$. Now let $x \to 0_+$. – Jean Marie Nov 01 '22 at 18:17
Thanks for your guys help – Nav Bhatthal Nov 01 '22 at 18:19
1

MSE $\tt MathJax$ Tutorial. – Felix Marin Nov 01 '22 at 18:35

0 Answers0