$f""(x)=$$limlimits_h o 0fracf"(x)-f"(x-h)h=limlimits_h o 0fracfracf(x+h)-f(x)h-fracf(x)-f(x-h)hh=limlimits_h o 0fracf(x+h)-2f(x)+f(x-h)h^2$

I would certainly really love input on this proof. The publication "Berkeley difficulties in brickandmortarphilly.comematics" solves the differently.

You are watching: Lim h approaches 0 f(x+h)-f(x)/h



Summing up$$f(x+h) - f(x) = h f"(x) + frach^22f""(x) + o(h^3)$$$$f(x-h) - f(x) = - h f"(x) + frach^22f""(x) + o(h^3)$$ girlfriend get$$f(x+h) + f(x-h) -2f(x) = h^2 f""(x) + o(h^3)$$which is tantamount to$$lim_h o 0 fracf(x+h) + f(x-h) -2f(x)h^2 = f""(x)$$

However i think this is the proof your book gives, due to the fact that this is quite standard.


Your proof is incorrect. You can sayeginalignf""(x)&=lim_h o 0fracf"(x)-f"(x-h)h\<10px>&=lim_h o 0dfraclimlimits_k o0dfracf(x+k)-f(x)k-limlimits_k o0dfracf(x-h+k)-f(x-h)khendalignbut at this point you"re nearly stuck. You could unify the two borders at the numerator and also work top top it, yet at the end you still have actually a twin limit.


Simply use L"Hospital ascendancy twice

$limlimits_h o 0dfracf(x+h)-2f(x)+f(x-h)h^2\=limlimits_h o 0dfracf"(x+h)-f"(x-h)2h\=limlimits_h o 0dfracf""(x+h)+f""(x-h)2\=f""(x)$


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