Limit of function doesn't exist Let $f$ be a real function and $x_0$ a real number. Suppose that $f(x)$ doesn't have a finite limit as $x \to x_0$. Does there necassary exist a sequence $z_1,z_2,...$ such that $\lim_{n \to \infty}z_n=x_0$, but $f(z_1), f(z_2),...$ doesn't have a finite limit?
 A: If $f(x)$ has a finite at $x_0$ then $\liminf_{x\to x_0} f(x)=\limsup_{x\to x_0} f(x)\in\mathbb{R}$. Hence if $f(x)$ does not have a finite at $x_0$ then we are in one of the following three cases:
1) $\limsup_{x\to x_0} f(x)=+\infty$;
2) $\liminf_{x\to x_0} f(x)=-\infty$;
3) $\limsup_{x\to x_0} f(x)=L \in \mathbb{R}$ and $\liminf_{x\to x_0} f(x)=l\in \mathbb{R}$ with $l<L$.
Now use the definition of $\limsup$ and $\liminf$ in order to find the required sequence $\{z_n\}_{n\geq 1}$:
1) there is a sequence $z_n\to x_0$ such that $f(z_n)=+\infty$;
2) there is a sequence $z_n\to x_0$ such that $f(z_n)=-\infty$;
3) for $\epsilon=(L-l)/3>0$, there is a sequence $u_n\to x_0$ such that $f(u_n)>L-\epsilon$ and there is a sequence $v_n\to x_0$ such that $f(v_n)<l+\epsilon$. Let $z_{2n}=u_n$ and $z_{2n-1}=v_n$, then $z_n\to x_0$ but $f(z_n)$ does not have a finite limit because 
$$f(z_{2n})-f(z_{2n-1})>(L-\epsilon)-(l+\epsilon)=(L-l)/3>0.$$
A: Hint: Try to build your sequence one by one, and by that making sure that it wont converge.
