Let $f$ be a differentiable function at $x=1$ such that $f(1)=1 , f'(1)=4$ I need to compute the following limit or prove it doesn't exist: $$ \lim_{x \to 1} \frac{1-f(x)}{x-1} $$

So I tried to figure out what is the limit of $\lim_{x \to 1}f(x)$

I started at the defenition of derivative: $$ \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0} => \lim_{x \to 1} \frac{f(x) - 1}{x-1} = 4 => \lim_{x \to 1} f(x) = 4x-3 $$

Now I'm not sure if I'm allowed to simply assign $4x -3 =\lim_{x \to 0} f(x)$ into the requested limit (first one) and just calculate it.. I'm new to the whole derivative thing so not sure exactly how it goes with limits I mean what exactly can and can't be done.

Please continue where I stopped and be as formal as you can.

Thank you

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    $\begingroup$ Hint: If $f(x)$ is differentiable at $1$ then, by definition, it is continuous there. $\endgroup$ – lulu Jan 14 '17 at 14:32
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    $\begingroup$ Note: your final expression does not make sense. The right hand. $4x-3$, is a function of $x$ but the left hand is not...the $x$ is just a dummy variable for the limit. $\endgroup$ – lulu Jan 14 '17 at 14:34
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    $\begingroup$ I cannot understand the close-vote at all $\endgroup$ – Peter Jan 14 '17 at 14:41
  • $\begingroup$ @lulu you right, thanks. $\endgroup$ – Noam Jan 14 '17 at 15:14

If $f$ is differentiable, then it is continuous, thus, $\lim_{x\to1}f(x)=f(1)$.

And by the definition of the derivative,


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    $\begingroup$ @Peter Maybe someone was thinking along the same lines? :-/ I have no idea. $\endgroup$ – Simply Beautiful Art Jan 14 '17 at 14:37
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    $\begingroup$ I just wondered how the upvoter could open the answer and notice that it is worth an upvote in such a short time ... $\endgroup$ – Peter Jan 14 '17 at 14:38
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    $\begingroup$ @Peter Same, way to fast IMO. Hm... $\endgroup$ – Simply Beautiful Art Jan 14 '17 at 14:38
  • $\begingroup$ Ok..not sure why you need to use the fact that f is continuous? $\endgroup$ – Noam Jan 14 '17 at 15:13
  • $\begingroup$ @Noam To point out that $\lim_{x\to1}f(x)=f(1)$, and you said So I tried to figure out what is the limit of $\lim_{x\to1}f(x)$ $\endgroup$ – Simply Beautiful Art Jan 14 '17 at 15:14

Because of $$4=f'(1)=\lim\limits_{x\to 1}\frac{f(x)-f(1)}{x-1}=\lim\limits_{x\to 1}\frac{f(x)-1}{x-1},$$ the limit in question equals $-4$.


You can also use L'Hopital's Rule:

$$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$

Provided that $f(x)/g(x)$ approaches some indeterminate form, such as $0/0$, as is the case in this problem (I'll leave it to you to verify).


Now plug in for $x$ and solve: $$\lim_{x\to1}\frac{0-f'(1)}{1}=\frac{-4}{1}=-4$$


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