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I don't know how to approach this exercise: Let $f$ be a real valued convex function that satisfies:

$$\lim_{x\to-\infty}f(x)=-\infty$$

prove that it must be $$\lim_{x\to+\infty}f(x)=+\infty$$

I know the definition of convexity and limits however I don't know how to prove this formally.

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    If $x_1 < x_2 < x_3$, then $$\frac{f(x_3) - f(x_2)}{x_3 - x_2} \geqslant \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$ for convex $f$. – 2017-01-02
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    Conceptual understanding of proofs is at least as important as the ability to follow or produce them mechanically. The answers given so far are all valid, but only vvnitram's gives a conceptual argument for why the result is true, and it assumes a little familiarity with convex functions. – 2017-01-03
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    Do you have a mental image of what convex functions look like? In particular, are you familiar with convex sets (the origin of the terminology)? Once you have a picture of the function $f$ in your head, the reason for the result you need to prove should be intuitively clear. Then you can formalize it using any of the equivalent arguments given below. – 2017-01-03

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Since $\lim_{x\rightarrow -\infty}f(x)=-\infty$, then there exists an $x_0<0$, s.t. $f(x_0)0. \end{equation} Now by convexity of $f$, we can show that for $x>0$, \begin{equation} \frac{f(x)-f(0)}{x-0}\geq \frac{f(0)-f(x_0)}{0-x_0}=k, \end{equation} i.e., $f(x)\geq f(0)+kx\rightarrow \infty$ as $x\rightarrow \infty$.

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Another hint: from $f\left(\frac{x_1+x_2}{2}\right) \leq \frac{f(x_1)+f(x_2)}{2}$, if we take $x_1=x$ and $x_2=-x$ then we have $2f(0) \leq f(x)+f(-x) \Rightarrow -f(x) + 2f(0) \leq f(-x)$. Taking the limit $$\lim_{x\rightarrow -\infty}\left(-f(x) + 2f(0)\right) \leq \lim_{x\rightarrow -\infty}f(-x)=\lim_{y\rightarrow \infty}f(y)$$ or $$\left(-\lim_{x\rightarrow -\infty}f(x)\right) + 2f(0) \leq\lim_{y\rightarrow \infty}f(y)$$

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Hint: if this lim is $-\infty$, $f$ must have a max, an absurd. If this lim is a real number $a$, the line $y=a$ is an asintota of $f$. Why this is an absurd?

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    The $\lim$ needs not exist, so you must first prove that it does in fact exist. – 2017-01-03