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To calculate the derivate of $f$ in $x=2$ I used cross multiplication but i can't explain it and it seems invalid, can u help? $$\lim_{x\to 4}{f(x)+7\over(x-4)}=-1.5$$ $$2f(x)+14=-3x+12$$ $$f(x) =-1.5x-1$$ So $$f'(x)=-1.5$$ Is it basically true?

Qurultay
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Boshra Alef
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    No because $f$ can be arbitrary outside any neighborhood of $4$ so we can‘t say something about $f'(2)$ – Maximilian Janisch Mar 14 '20 at 17:22
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    Does it not directly follow from the limit that $f'(x)=-3/2$? Besides $f(x)$ for which this holds is not unique. – Paras Khosla Mar 14 '20 at 17:22
  • Is there any helpful context what $\ f(x)\ $ is ? For example, do we know that it is a polynomial ? – Peter Mar 14 '20 at 17:26
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    @maximilianjanisch you're right but basically is it a valid way to obtain the function's formula when the limit is 0/0? I mean the cross multiplication. – Boshra Alef Mar 14 '20 at 17:28
  • @Peter : You have the implication in the invalid direction. The correct implication: If $f'(4)$ exists, the given limit equation shows that $f'(4) = -1.5$. However, nothing here forces $f'(4)$ to exist. – Eric Towers Mar 14 '20 at 17:33
  • @EricTowers You are right. Anyway, I guess that $f(x)$ here is a sufficiently smooth function. – Peter Mar 14 '20 at 17:34

4 Answers4

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No, it is not correct. You cannot jump from$$\lim_{x\to4}\frac{f(x)+7}{x-4}=-\frac32\text{ to }f(x)+7=-\frac32x+6.$$The limit cannot just vanish like that.

Since $\lim_{x\to0}x-4=0$ and since the limit $\lim_{x\to4}\frac{f(x)+7}{x-4}$ exists (in $\mathbb R$), the limit $\lim_{x\to4}f(x)+7$ must be equal to $0$. So (assuming that $f$ is continuous at $4$), $f(4)=-7$. So$$f'(4)=\lim_{x\to4}\frac{f(x)-f(4)}{x-4}=-\frac32.$$

José Carlos Santos
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  • You said the limit can not vanish, can you explain it more and say why? Thanks in advance – Boshra Alef Mar 14 '20 at 17:32
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    Because if it could vanish, you would conclude that $f$ could only be the maps $f(x)=-\frac32x-\frac12$. But that's not true. For instance, $f$ can also be the function $f(x)=-\frac32x-\frac12+(x-4)^2$. – José Carlos Santos Mar 14 '20 at 17:39
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If you have used cross multiplication here you only know the behavior of $f(x)$ in the neighborhood of $x=4$, if $f(x)$ continuous it gives correct value of $f(4)$. You also get the correct value of $f'(4)$. By doing this cross multiplication you cannot get function at any value other than $x=4$.

Z Ahmed
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No, you can't remove the "$\lim_{x \to 4}$" part like that. Neither can you know anything about the function $f$ except that $\lim_{x \to 4} f'(x) = -\frac{3}{2}$ (under two hypothesis, see below). Here's how :

Assume (1) $f(4) = -7$ (otherwise you can't use l'Hôpital rule). Since $\lim_{x \to 4} \frac{f(x) + 7}{x - 4}$ is of the undetermined form $\frac{0}{0}$, one can use L'Hôpital's rule to compute this limit.
The derivative of the numerator is $f'(x)$. The derivative of the numerator is $1$. Hence :

$$ -\frac{3}{2} = \lim_{x \to 4} \frac{f(x) + 7}{x - 4} = \lim_{x \to 4} \frac{f'(x)}{1} $$

Now, if $f'$ is continuous, this implies $f'(4) = - \frac{3}{2}$.

Note about "assumption" (1)
Since $\lim_{x \to 4} \frac{f(x) + 7}{x - 4} = -\frac{3}{2}$ although the denominator tends to $0$, the numerator has to tend to $0$, for otherwise we wouldn't have a finite limit for the quotient. Assumption (1) comes for free.

Olivier Roche
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  • This is , to my opinion, the best answer because it gives the sufficient condition (which probably is supposed to hold here) that we can actually calculate $f'(4)$ – Peter Mar 14 '20 at 17:36
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    LHospital works on $0/0$ and not on $l/0$. I guess you have a typo there. Another issue is regarding continuity of derivative $f'$. That's too much to assume. It is best to assume of $f$ at $4$ and then one concludes $f'(4)=-3/2$. Also note that derivative evaluation can't be based on LHospital but rather it works in reverse. You use idea of derivatives while applying LHospital. – Paramanand Singh Mar 14 '20 at 18:50
  • @ParamanandSingh You are right, I edit this! – Olivier Roche Mar 14 '20 at 19:48
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Your approach indicates that you are still in Algebra mode which involves mostly symbol manipulation using $+, - \times, /$. When dealing with limits (and calculus in general) one must leave that mode and instead use theorems meant to deal with them.

You are given that $$\lim_{x\to 4}\frac{f(x)+7}{x-4}=-\frac{3}{2}$$ and the above does not give us any clue about $f$ apart from its local behavior near point $4$. In particular one can't even conclude what $f(4)$ is. This is simply because different functions can have same limit at a point. In other words limit at a point does not identify a function uniquely even in a neighborhood of the point under consideration.

And by now you must have understood that the given information can't be used to conclude anything about derivative of $f$ at $2$.

The given limit however does ensure that $\lim_{x\to 4}f(x)=-7$. How? Well, by using limit laws we have $$\lim_{x\to 4}f(x)=\lim_{x\to 4}\frac{f(x)+7}{x-4}\cdot(x-4)-7=(-3/2)\cdot 0-7=-7$$ You can see that the first step here is pure algebra but the next step is not. So you don't leave the algebra mode altogether but rather use it only till the point it is needed and no further.

Next if we are given additional hypothesis that $f$ is continuous at $4$ then we can conclude that $f(4)=\lim_{x\to 4}f(x)=-7$.

If on the other hand we assume that the question is modified to add the hypothesis of continuity of $f$ at $4$ and we are asked to evaluate $f'(4)$ then the given limit in question implies almost immediately that $f'(4)=-3/2$.

There is another aspect which you must also learn here. Typical evaluation of derivatives involves a scenario where the function is specified by a formula containing the variable $x$ (or whatever symbol one prefers). In such a case one uses various rules of differentiation and a table of derivatives for well known functions (technical word is elementary functions). This process is almost mechanical and sort of looks like fancy algebraic manipulation.

If the function is not specified by a simple formula then one has to go to the basics and use the definition of derivative as a limit and evaluate desired limit via limit laws.

Paramanand Singh
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