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Given that $x^3[f(x+1)-f(x-1)]=1$, determine $\lim_{x\rightarrow \infty}(f(x)-f(x-1))$ explicitly.

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I think you need continuity here. – TonyK Jan 19 '11 at 17:43
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If $g(x)=f(x)-f(x-1)$, then $g(x+1)+g(x)=1/x^3$. It follows that if your limit exists, then its value is zero. But there is no reason for it to exists: you can construct pretty wild solutions by starting from an arbitrary function on an interval of length $2$ and using the equation to continue it. – Mariano Suárez-Alvarez♦ Jan 19 '11 at 17:45

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if $\lim_{x\to\infty}(f(x)-f(x-1))$ exists and it's equal to $l$, then also $\lim_{x\to\infty}(f(x+1)-f(x))=l$. So $$ 2l=\lim_{x\to\infty}( f(x+1)-f(x-1))=\lim_{x\to\infty}\frac{1}{x^3}=0, $$ hence $l=0$.

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