# Evaluating the limit (inequation)

Could you help me out how to calculate the limit $$\lim\limits_{j \rightarrow 0}{\frac{f(j(x_1-x_0)+x_0)-f(x_0)} {j(x_1-x_0)}\cdot(x_1-x_0) \leq f(x_1)-f(x_0)}\;, j\in[0,1]$$ Thanks!

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What exactly do you mean by "what happens to this inequation [I assume you mean inequality] when building the limit?" Also, what have you tried in order to solve the problem? – Gyu Eun Lee Jan 12 '13 at 23:29
Are you missing a $j$ somewhere in the numerator? You currently have a constant term $f(x_0) \cdot (x_1 - x_0)$. – Calvin Lin Jan 12 '13 at 23:42
What is j, a function? – lee Jan 13 '13 at 1:50
My bad, j $\in [0,1]$ – Josh Jan 13 '13 at 1:58
The question makes even less sense now...you have an inequality yet you want to calculate the limit of the LHS? Do you mean you want to prove the inequality? And again, what have you tried so far? (For all we know you may be 90% to the answer!) – Gyu Eun Lee Jan 13 '13 at 2:21

By definition $$f'=\lim\limits_{h \rightarrow 0}{\frac{f(h+x_0)-f(x_0)} {h}}\;$$ when $h=(x-x_0).$ We see that as j approaches 0 $$\lim\limits_{j \rightarrow 0}{\frac{f(j(x_1-x_0)+x_0)-f(x_0)} {j(x_1-x_0)}}=f'(x_0),$$ so when passing to the limit in the inequality $$\lim\limits_{j \rightarrow 0}{\frac{f(j(x_1-x_0)+x_0)-f(x_0)} {j(x_1-x_0)}\cdot(x_1-x_0) \leq f(x_1)-f(x_0)}\;$$ we conclude that $$f'(x_0)\cdot(x_1-x_0) \leq f(x_1)-f(x_0).$$
Well, after ${j \rightarrow 0}$ you see that $f'(x_0)\cdot(x_1-x_0) \leq f(x_1)-f(x_0).$ – Josh Jan 13 '13 at 16:15
OK. A more readable version would be: "Passing to the limit in the inequality $\lim\limits_{j \rightarrow 0}{\frac{f(j(x_1-x_0)+x_0)-f(x_0)} {j(x_1-x_0)}\cdot(x_1-x_0) \leq f(x_1)-f(x_0)}$, we conclude that $f'(x_0)\cdot(x_1-x_0) \leq f(x_1)-f(x_0)$." – user53153 Jan 13 '13 at 16:19
Well, the answer is readable now. I suppose you were given the inequality $${\frac{f(j(x_1-x_0)+x_0)-f(x_0)} {j(x_1-x_0)}\cdot(x_1-x_0) \leq f(x_1)-f(x_0)}$$ somewhere? You did not state it in the question above. In fact, your question asked the readers to prove a certain inequality for $f$ without giving them any information about $f$. – user53153 Jan 13 '13 at 16:39