1
$\begingroup$

I have the next $\lim_{x \to \frac{\pi}{4}} \frac{f(x)-f(\frac{\pi}{4})}{x-\frac{\pi}{4}}$ when the function is $$f(x)=\frac{x+\sin(x)}{\tan(x)}.$$ I don't know how to even start. I am sorry that this is short but I really don't know.

$\endgroup$
6
  • 3
    $\begingroup$ Take a look at that limit. Look at the form of it. Does it look, in any way, shape or form, familiar? $\endgroup$
    – Arthur
    Commented Feb 7, 2021 at 11:57
  • 3
    $\begingroup$ What is the definition of the derivative of a function? Any relation with your limit? $\endgroup$
    – Robert Z
    Commented Feb 7, 2021 at 11:57
  • $\begingroup$ @Arthur I can't see, can you show me? $\endgroup$
    – acid1302
    Commented Feb 7, 2021 at 11:59
  • 2
    $\begingroup$ Give it a try!! What is the definition of the derivative of a function? $\endgroup$
    – Robert Z
    Commented Feb 7, 2021 at 12:00
  • 1
    $\begingroup$ You have a BIG hint. Read en.wikipedia.org/wiki/Derivative#Rigorous_definition Now it is your turn! $\endgroup$
    – Robert Z
    Commented Feb 7, 2021 at 12:06

1 Answer 1

4
$\begingroup$

$f$ is said to be differentiable at $c$, if there exists a real number $\alpha$ such that $$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\alpha.$$ Now see your question.

Find $f'$ and then put $x=\frac{π}{4}.$


Note that $\frac{d}{dx}f=f'.$ If $$f(x)=\frac{h(x)}{g(x)}$$, then $$f'(x)=\frac{g(x)\frac{d}{dx}h(x)-h(x)\frac{d}{dx}g(x)}{(g(x))^2}.$$

$\endgroup$
1
  • $\begingroup$ so is it $1-\frac{\pi +\sqrt{2}}{2}$ $\endgroup$
    – acid1302
    Commented Feb 7, 2021 at 12:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .