Contradiction in derivatives as linear approximations

Question

From the definition of a derivative, we have that $$f'(a) = \lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$

or $$\lim\limits_{x\to a}f'(x) = \lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$

This leads me to believe we can write $$\lim\limits_{x\to a}(f(a)+(x-a)f'(x))=\lim\limits_{x\to a}f(x)$$

Which to me, implies that derivative allow us to obtain a good approximation for a function near $a$, and that this approximation will become increasingly accurate as $x\to a$.

However, couldn't we just as easily write

$$\lim\limits_{x\to a}(f(a)+2(x-a)f'(x))=\lim\limits_{x\to a}f(x)$$

since $$\lim\limits_{x\to a}(f(a)+(x-a)f'(x))=f(a)+\lim\limits_{x\to a}((x-a)f'(x))$$

and $$\lim\limits_{x\to a}((x-a)f'(a)) = 0 = \lim\limits_{x\to a}(2(x-a)f'(a))$$ provided that $\lim\limits_{x\to a}f'(a)$ exists.

Wouldn't this imply instead that $\lim\limits_{x\to a}(f(a)+2(x-a)f'(x))$ can be used to provide good approximations for the function near $a$? In fact, couldn't we replace $2$ with any other constant or function which has a limit at $a$?

Using such approximations, however, would obviously produce incorrect results in proofs such as those for the chain and product rules. So how can this contradiction be dealt with? Why is using $f'(x)$ simply more correct than using $2f'(x)$, even when the math doesn't necessarily seem to be demonstrating this?

Answer 1

Note that when $f$ is continuously differentiable at $a$, then $$\lim\limits_{x\to a}(f(a)+(x-a)f'(x))=\lim\limits_{x\to a}f(x)$$ is certainly true, but it is also true that

$$\lim\limits_{x\to a}(f(a)+g(x))=\lim\limits_{x\to a}f(x)$$

for any function $g$ such that $\lim_{x\to a}g(x)=0.$ Clearly, using the above limit equations as a basis for approximation is not going to get you anywhere meaningful.

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Instead, without any limits, we can simply see from the definition of the derivative that when $f$ is differentiable at $a$, we can approximate the value of $f$ at $x$ as

$$f(x)\approx f(a)+(x-a)f'(a).\quad (1)$$

In fact, the expression in $(1)$ is the first order Taylor polynomial of $f$ around $a$. This may not be a "good" approximation; how "good" it is depends on your error tolerance, and there are known results on the bound of the error of such polynomials. See here.

Answer 2

$$\lim\limits_{x\to a}f'(x) = \lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$

This claim assumes that $f'$ is necessarily continuous (what we might write as $f \in \mathcal{C}^1(D)$ for the appropriate domain $D$). This is not always the case; some examples are here, such as the function

$$f(x) := \begin{cases} x^2 \sin(1/x), & x \ne 0 \\ 0, & x = 0 \end{cases}$$

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$$\lim\limits_{x\to a}(f(a)+(x-a)f'(x))=\lim\limits_{x\to a}f(x)$$

Claiming this, further, makes certain assumptions about limit laws that need not hold. In particular, more or less what you're doing is claiming

$$\lim_{x \to a} \frac{f(x)-f(a)}{x-a} = \frac{\displaystyle \lim_{x \to a} f(x)-f(a)}{\displaystyle \lim_{x \to a}x-a} \tag{1}$$

and then multiplying by the denominator to get

$$\left( \lim_{x \to a} f'(x) \right) \left( \lim_{x \to a} (x-a) \right) = \lim_{x \to a} \Big( f(x) - f(a) \Big) \tag{2}$$

and then splitting the latter limit to get

$$\left( \lim_{x \to a} f'(x) \right) \left( \lim_{x \to a} (x-a) \right) = - f(a) + \lim_{x \to a} f(x) \tag{3}$$

and then adding $f(a)$ to both sides,

$$f(a)+ \left( \lim_{x \to a} f'(x) \right) \left( \lim_{x \to a} (x-a) \right) = \lim_{x \to a} f(x) \tag{4}$$

and then recombining the left hand side all under the same limit:

$$ \lim_{x \to a} \Big( f(a) + f'(x) (x-a) \Big) = \lim_{x \to a} f(x) \tag{5}$$

Quite a few rules for limits were broken along the way.

• In $(1)$, we can only claim that $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\displaystyle \lim_{x\to a} f(x)}{\displaystyle \lim_{x\to a} g(x)}$$ when all three limits involved exist (as finite numbers, to be clear: $\pm \infty$ is not a case where we say a limit exists here), and the bottom limit of the fraction is nonzero. A problem arises: the bottom limit goes to zero in $(1)$.

• In $(2)$, you just multiplied both sides by $0$ as a result; $(2)$ is identical to saying $0=0$, so no actual information of note may be derived.

• In $(3)$, I would be hesitant to say you can split the limit up like that. Like the above rule, for any $\alpha,\beta \in \mathbb{R}$, we can only say $$ \lim_{x \to a} \Big( \alpha f(x) + \beta g(x) \Big) = \alpha \lim_{x \to a} f(x) + \beta \lim_{x \to a} g(x)$$ when all three limits exist. Does the limit as $x \to a$ of $f(x)$ necessarily exist? One should do more work to justify this claim.

• $(5)$ has many of the same concerns as all of the previous rules mentioned (aside from a product equivalent to the quotient law), just applied in reverse.

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Granted, for a restricted class of functions, your equality does hold, and we have the more general observation mentioned in Golden_Ratio's answer that

$$\lim_{x \to a} \Big( f(a) + g(x) \Big) = \lim_{x \to a} f(x)$$

when $g(x) \xrightarrow{x\to a} 0$, so one begs the question: why is the formula

$$f'(x) \approx f(a) + f'(a) (x-a)$$

considered a good approximation?

One can certainly handwave some details. For $x$ very much near $a$, if

$$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$$

is expected to hold, then for $x$ very near $a$ (say, $x = a +\varepsilon$ for some $\varepsilon$ very very small with $|\varepsilon|>0$), the limit and derivative should be about the same. That is,

$$f'(a) \approx \frac{f(a+\varepsilon) - f(a)}{a +\varepsilon - a} = \frac{f(a+\varepsilon)-f(a)}{\varepsilon}$$

But then rearranging,

$$f(a+\varepsilon) \approx \varepsilon f'(a) + f(a)$$

and since $\varepsilon = x-a$, the desired approximation arises. Of course, when concerned with rigor, one should avoid such hand-waving.

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A past MSE post somewhat handles why the tangent-line approximation is the "best" linear approximation (since your scalings of $f'(x)$ still render the approximation linear) -- it is the only linear function (in the sense of a form $y=mx+b$) whose error tends to $0$ faster than $x-a$ tends to $0$.

Also of interest may be Taylor's theorem, particularly results tied to the remainder term and bounds on it.

Answer 3

The point of the derivative as a linear aproximation of a function $f$ at a point $a$ is that the derivative is not just a linear aproximation, is the best linear aproximation, but, given a function $f : \mathbb{R}\rightarrow \mathbb{R}$ and $a\in \mathbb{R}$, intuitively, what means that a liner aproximation $L(x) = m(x-a) + f(a)$ of $f$ at $a$ is the best? Notice that is not enought if we say that $L$ is the best linear aproximation if: $\lim_{x\to a}L(x) = f(a)$, since every function of the form $m(x-a) + f(a)$ satisfies this, so what conditions are reasonable to say that $L$ is the best linear aproximation? That $L$ is the best linear aproximation means that you don't need to stay very close to $a$ to aproximate very accurately $f$, i.e. given a $\epsilon > 0$ (very very small), you can find a $\delta > 0$ (not too small) such that for $x\in \mathbb{R}$ with $|x-a|<\delta$ we get that $|L(x) - f(x)|<\epsilon$, in other words, $L(x)-f(x)$ goes to $0$ faster than $x-a$ does. This reasoning lead us to say that $L$ is the best linear aproximation if

$$\lim_{x\to a}\frac{f(x)-L(x)}{x-a} = 0$$

and this reasoning gives us the correct answer since if the derivative exist at a point, you can rewrite the limit

$$f'(a) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a} $$

as

$$\lim_{x\to a}\frac{f(x)-[f(a) + (x-a)f'(a)]}{x-a} = 0$$

and is easy to prove the following:

Let $f : A\subseteq\mathbb{R}\rightarrow\mathbb{R}$ be a function and $a\in A'$. $f$ is differentiable at $a$ if and only if there exists $m\in\mathbb{R}$ such that $$\lim_{x\to a}\frac{f(x)-[f(a) + m(x-a)]}{x-a} = 0$$ in any case, we get that $f'(a) = m$.