2

If $f$ is differentiable and a strictly increasing function, then what is the following?

$$\lim_{x\to0}\frac{f(x^2) - f(x)}{f(x)-f(0)}$$

Thank you.

zibzabpuzzless
  • 455
  • 1
    If $f'(0)\neq0$, then: $$\lim_{x\to0}\frac{f(x^2) - f(x)}{f(x)-f(0)}=\lim_{x\to 0} \frac{f(x^2)-f(0)+f(0)-f(x)}{x} \cdot \frac{x-0}{f(x)-f(0)}=\lim_{x\to 0}\left(x \cdot \frac{f(x^2)-f(0)}{x^2-0}-\frac{f(x)-f(0)}{x-0}\right)\cdot \frac{x-0}{f(x)-f(0)} =(0 \cdot f'(0)-f'(0))\cdot \frac{1}{f'(0)}=-1.$$ – npatrat May 23 '15 at 13:08
  • This has been very helpful, thank you – zibzabpuzzless May 23 '15 at 13:15
  • It can happen that $f'(0)=0$ exists, but $$\lim_{x\to0}\frac{f(x^2) - f(x)}{f(x)-f(0)}$$ does not exist. – GEdgar May 23 '15 at 16:11

2 Answers2

3

You could even do a bit more if you practice Taylor series. Built at $x=0$, you have $$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+\frac{1}{6} f'''(0) x^3+O\left(x^4\right)$$ Doing the same (using the chain rule) $$f(x^2)=f(0)+x^2 f'(0)+O\left(x^4\right)$$ So, $$\frac{f(x^2) - f(x)}{f(x)-f(0)}=\frac{-x f'(0)+x^2 \left(f'(0)-\frac{f''(0)}{2}\right)-\frac{1}{6} f'''(0) x^3+O\left(x^4\right)}{x f'(0)+\frac{1}{2} x^2 f''(0)+\frac{1}{6} f'''(0) x^3+O\left(x^4\right)}$$ Now, performing the long division (provided that $f'(0)\neq 0$) $$\frac{f(x^2) - f(x)}{f(x)-f(0)}=-1+x-\frac{ f''(0)}{2 f'(0)}x^2+O\left(x^3\right)$$ which shows not only the limit but also how it is approached.

Edit

Using the same approach, you could show that, for any $n \geq 2$, $$\frac{f(x^n) - f(x^{n-1})}{f(x^{n-1})-f(0)}=-1+x-\frac{ f''(0)}{2 f'(0)}x^n+\cdots$$

Claude Leibovici
  • 260,315
0

$\lim_{x \arrow 0} \frac{f'(x^2)(2x)-f'(x)}{f'(x)-0}$ using L'hospital rule $=\frac{f'(0)(0)-f'(0)}{f'(0)-0}$ $=\frac{-f'(0)}{f'(0)}=-1$

user157012
  • 91
  • 1
  • 5