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Q. Find the derivative of the $sin\,x$ at $x = 0$

Limits and Derivatives

Solution:

Let $f(x) = sin\, x$. Then
$f'(0)=\displaystyle \lim_{h \to 0}$ $\frac{f \left(0+h\right)-f \left(0\right)}{h}$
$=\displaystyle \lim_{h \to 0}$ $\frac{sin\left(0+h\right)-sin\left(0\right)}{h}$
$=\displaystyle \lim_{h \to 0}$ $\frac{sin\,h}{h}=1$