2
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Hint: If $f(x)$ is differentiable at $1$ then, by definition, it is continuous there. $\endgroup$ – lulu Jan 14 '17 at 14:32
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
6
$\begingroup$

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

And by the definition of the derivative,

$$f'(1)=\lim_{x\to1}\frac{f(x)-f(1)}{x-1}$$

$\endgroup$
  • 1
    $\begingroup$ @Peter Maybe someone was thinking along the same lines? :-/ I have no idea. $\endgroup$ – Simply Beautiful Art Jan 14 '17 at 14:37
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
7
$\begingroup$

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 $-4$.

$\endgroup$
3
$\begingroup$

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).

$$\lim_{x\to1}\frac{1-f(x)}{x-1}=\lim_{x\to1}\frac{0-f'(x)}{1-0}$$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.