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friends, there is a formula derivation I saw in a book: $$ {\displaystyle \lim_{ \triangle x \rightarrow 0}{\frac{ \triangle y}{ \triangle x}} } =f'(x_0) \Rightarrow {\displaystyle \lim_{\triangle x \rightarrow 0}{\frac{\triangle y}{\triangle x}}} - f'(x_0)=0 \Rightarrow \frac{\triangle y}{\triangle x} - f'(x_0)=a, {\displaystyle \lim_{\triangle x \rightarrow0}{a} = 0} \Rightarrow \triangle y = f'(x_0)\triangle x + a\triangle x $$

how to understand this step?

$$ {\displaystyle \lim_{\triangle x \rightarrow 0}{\frac{\triangle y}{\triangle x}}} - f'(x_0)=0 \Rightarrow \frac{\triangle y}{\triangle x} - f'(x_0)=a, {\displaystyle \lim_{\triangle x \rightarrow0}{a} = 0} $$


EDIT-01

Thanks for Gerry Myerson's comment.

as you say,

name $\frac{\triangle y}{\triangle x}-f'(x_0)$ as $a$,

but, the previous step is $ ({\displaystyle \lim_{x \rightarrow 0}{\frac{\triangle y}{\triangle x}}}) - (f'(x_0)) = 0 $, not ${\displaystyle \lim_{x \rightarrow 0}{(\frac{\triangle y}{\triangle x}} -f'(x_0)) } = 0$.

there I don't understand.

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    $\begingroup$ I don't know where the $x_0$ in $\Delta x_0\to0$ comes from, why it's not just $\Delta x\to0$ as it is elsewhere in the formulas. But anyway all that's going on in the step you ask about is that the quantity $(\Delta y/\Delta x)-f'(x_0)$ is being given the name $a$, and then from the left side of the implication it follows that $\lim_{\Delta x\to0}a=0$. $\endgroup$ – Gerry Myerson Dec 4 '20 at 21:29
  • $\begingroup$ Understood now? $\endgroup$ – Gerry Myerson Dec 6 '20 at 0:22
  • $\begingroup$ It's a typo, now I fixed it. My mistake. Thank you for pointing it out. $\endgroup$ – 244boy Dec 6 '20 at 1:40
  • $\begingroup$ OK...but, do you now understand the step that was bothering you, that I explained? $\endgroup$ – Gerry Myerson Dec 6 '20 at 4:19
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    $\begingroup$ In the edit, you have $\lim_{x\to0}$ instead of $\lim_{\Delta x\to0}$, but I'll assume that's just a typo. Anyway, $f'(x_0)$ is a constant, not dependent on $\Delta x$ (nor on $x$), so $$\lim_{\Delta x\to0}\left({\Delta y\over\Delta x}-f'(x_0)\right)=\left(\lim_{\Delta x\to0}{\Delta y\over\Delta x}\right)-\lim_{\Delta x\to0}f'(x_0)=\left(\lim_{\Delta x\to0}{\Delta y\over\Delta x}\right)-f'(x_0)$$ $\endgroup$ – Gerry Myerson Dec 6 '20 at 21:03

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