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As I was a student, I got the following problem:

Let $f$ be a mapping from $[a,b]$ to $\mathbb{R}$ satisfying $$(1)\quad f((1-\lambda)x+\lambda y)\leq (1-\lambda )f(x)+\lambda f(y)$$ for all $x,y\in [a,b]$ and $\lambda\in [0,1]$. Suppose $f$ is differentiable on $]a,b[$. Show that if $f$ is continuous and $f'$ increasing then $(1)$ holds and vice versa.

As a part of the solution, I fixed $y$, assumed $x<y$, and denoted $z:=(1-\lambda)x+\lambda y$ where $\lambda\ne 0,1$. Then we have $x\leq z$. I also proved that $$\frac{f(z)-f(x)}{z-x}\leq \frac{f(y)-f(z)}{y-z}$$

Now I thought that $$\lim_{z\to x}\frac{f(z)-f(x)}{z-x}\leq \lim_{z\to x}\frac{f(y)-f(z)}{y-z}.$$

My advisor said it is wrong and it should be

$$\lim_{z\to x+}\frac{f(z)-f(x)}{z-x}\leq \lim_{z\to x+}\frac{f(y)-f(z)}{y-z}.$$

I never got a proper reasoning why the formula I wrote is wrong. I assumed that $x\leq z$ so I thought $\lim_{z\to x}$ means $\lim_{z\to x+}$ as $z$ can't approach $x$ from left. Could anyone explain me why limit does not mean one sided limit if I can't compute the two-sided limit?

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As you said $z\geq x$, so $z$ cannot approach $x$ from the left. The notation $\lim_{z\to x}$ implies that $z$ could be approaching from either side.

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