Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.