What is the Limit Definition of a Tough Derivative?

member 428835
hey pf!

can you help me with this $$\lim_{h \to 0} \frac{f(x+3h^2) - f(x-h^2)}{2h^2}$$

i know the definition and have tried several substitutions, but no help. anyone have any ideas?
 
Physics news on Phys.org
nevermind, lopitals rule did the trick
 
hint 1
$$0=\mathrm{f}(x)-\mathrm{f}(x)$$
hint 2
$$\lim_{h \to 0} \frac{f(x+3h^2) - f(x-h^2)}{2h^2}=\lim_{h \to 0}\left[\frac{3}{2}\frac{\mathrm{f}(x+3h^2)-\mathrm{f}(x)}{3h^2}+\frac{1}{2}\frac{\mathrm{f}(x-h^2)-\mathrm{f}(x)}{-h^2}\right]$$
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top