I am trying to prove to myself that, starting with the definition of the derivative

$$f'(x)=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}$$

[Note: I wrote the above mistakenly, as pointed out by a comment. Since I think this "typo" was really a big source of my confusion, I'm leaving it as is, to see why I ever asked the question. Of course, $f'(x)$ should be $f'(x_0)$.]

this is equivalent to

$$f'(x)=\lim_{t\rightarrow 0}\frac{f(x+t)-f(x)}{t}$$

I know that I may define $t=x-x_0$ and then the difference quotient becomes


Expanding the definition of the limit in the definition of the derivative we have, for any $\varepsilon>0$ there is some $\delta>0$ such that

$$|x-x_0|<\delta \Rightarrow \left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x)\right|<\varepsilon$$

This is equivalent to

$$|t|<\delta \Rightarrow \left|\frac{f(x_0+t)-f(x_0)}{t}-f'(x)\right|<\varepsilon$$

So my problem is that what the second definition of the limit seems to say is not what I want it to say. It seems to say:

$$\lim_{t\rightarrow 0}\frac{f(x_0+t)-f(x_0)}{t} = f'(x)$$

Whereas it seems to me it should say:

$$\lim_{t\rightarrow 0}\frac{f(x_0+t)-f(x_0)}{t} = f'(x_0)$$

Now I'm also somewhat perplexed by what appears to me to be a shift from $x_0$ to $x$ in the typical way that this conversion is done. Typically, in a simpler calculus context, people would just make the fast argument (without reference to the analytic definition of the limit):

$$\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0} = \lim_{t\rightarrow 0}\frac{f(x_0+t)-f(x_0)}{t} = \lim_{t\rightarrow 0}\frac{f(x+t)-f(x)}{t}$$

This disturbs me a little because it shifts without explanation from $x$ being a variable and $x_0$ being a fixed parameter, to $x_0$ being a variable. However, even if I were to understand this switch and become comfortable with it, it seems to me that you cannot perform the same switch inside the definitions of the limit, as you would be inconsistently replacing some instances of $x$ with $x_0$, even though this value retains its earlier meaning elsewhere in the expression.

  • 2
    $\begingroup$ In your very first equation of the post you should have $f'(x_0)$ on the LHS. $\endgroup$ – MisterRiemann May 19 at 14:36

I believe you are correct for the most part, except that the definitions read: $$ f'(x_0)=\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$$ $$ f'(x_0)=\lim_{t\to 0} \frac{f(x_0+t)-f(x_0)}{t}.$$ So, the difficulty you are concerned about is never encountered. One simply performs the "change of variables" $t=x-x_0$ and notices that as $x\to x_0$, $t\to 0$ and conversely.


It doesn't make sense to define$$f'(x)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0},$$since the expression after the $=$ sign depends only upon $x_0$, So, the correct possible definitions are$$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\text{ and }f'(x_0)=\lim_{t\to0}\frac{f(x_0+t)-f(x_0)}t,$$which are clearly equivalent (given $\varepsilon>0$, if a $\delta>0$ works for one of them, then it also works for the other one).


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