Here is the link (problem $3$)

I could not understand this step

$$(x-x_0)<\frac{f'(\xi)}{f'(x)}(x-x_0)=\frac{f(x)-f(x_0)}{f'(x)}<0$$ Now taking the limit $$-x_0\leq\lim_{x\to0^+}\frac{f(x)\color{red}{-f(x_0)}}{f'(x)}<0$$ Where did this go (the part in red)?

If you say the $x_0$ is fixed and $f'$ is tends infinity therefore that thing becomes zero but then in the LHS we are also kinda of taking $x_0$ very close to zero (in a way $x_0$ is also tends to zero) so when we are tending $x_0$ to something then how come it is constant. I can't digest this?

  • 1
    $\begingroup$ I may be wrong, but you're taking the limit $x\rightarrow 0^{+}$, thus in the lhs $x_0$, which is fixed, remains $x_0$, the $x$ vanishes, and in the rhs $\frac{f(x_0)}{f'(x)}$ goes to 0 as it is the limit of a costant over something going to $-\infty$ $\endgroup$
    – qwertyguy
    Feb 11, 2020 at 15:18
  • $\begingroup$ that what i am thinking but then to prove that the limit is zero we have to take Xo very very close to zero, you might take very small value of Xo and then fix it but it doesnt matter how small Xo is limit wont be zero till we tend Xo to zero $\endgroup$ Feb 11, 2020 at 15:38

1 Answer 1


First we fix $x_0$ (no need to be however close to $0$), thus the missing $f(x_0)$ is also a fixed finite value.
But, we let $x$ tend to $0$ (from the right), and as $f'(x)\to -\infty$, we get $$\lim_{x\to 0+}\frac{f(x_0)}{f'(x)}=0$$ that's why it disappears from the next line.

Then we can either apply another limiting argument for $x_0$, or, as written, just observe that the resulting formula holds for all $x_0>0$ and conclude by this.


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