Mathematics Asked on January 4, 2021
In this post a user claims
begin{align*}
f'(x)
&=lim_{hto 0}frac{f(x+h)-f(x)}{h}\
&=lim_{hto 0}frac{sin 3(x+h)-sin(3x)}{h}\
&=lim_{hto 0}frac{2cos (3x+frac32h)sin frac32h}{h}\
&=lim_{hto 0}frac{2cos (3x+frac32h)frac32h}{h}\
&=lim_{hto 0}3cos (3x+frac32h)\
&=3cos(3x)
end{align*}
How did he get $frac 32h$ from $sin(frac 32h)$?
I don’t understand.
Small angle approximation, probably. $sin(x) approx x$ for $x$ small.
More rigorously, asymptotic equivalence. Note the well-known limit
$$lim_{x to 0} frac{sin(x)}{x} = 1$$
This gives the asymptotic equivalence $sin(x) sim x$ as $x to 0$. This means, within some circumstances, you can interchange $x$ and $sin(x)$ in limits where $x to 0$. (In this case, $x=3h/2$.)
An alternative angle of approach as pointed out by player3236 in the comments is that they could have multiplied to get a cancellation as so:
$$require{cancel}frac{2cos (3x+frac32h) cancel{sin frac32h}}{h} cdot frac{frac 3 2 h}{cancel{sin frac 3 2 h}} = frac{2cos (3x+frac32h)frac32h}{h}$$
Correct answer by Eevee Trainer on January 4, 2021
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