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Let $f(x)$ be a real valued function with its domain in $mathbb{R}$. Is there an example of $f(x)$ such that
$$lim_{xrightarrow x_0}f(x)=infty$$
for all $x_0$ in some interval?

Assuming that "domain in $Bbb R$" means $f:Bbb RtoBbb R$: No. Say $f:[a,b]toBbb R$. For $ninBbb Z$ let $$E_n={xin[a,b]:f(x)le n}.$$Then $$[a,b]=bigcup_{ninBbb Z} E_n.$$Since $[a,b]$ is uncountable this shows that there exists $n$ such that $E_n$ is infinite. So $E_n$ has an accumulation point $x_0in[a,b]$, and hence $f$ does not tend to infinty at $x_0$.

Short answer: Yes.

While David C. Ullrich has given what seems to be a good answer, there's a fatal flaw. It assumes that $f$ is defined everywhere in some interval. The question, however, only says that the domain is in $Bbb{R}$, and makes no other assertions.

The function:

In fact, there is a function $f$ which is defined on the rationals that has the property you want. In fact, it has the property that for all $x_0inBbb{R}$ it has $$lim_{xto x_0} f(x)= infty.$$

Define $f:Bbb{Q}to Bbb{R}$ by $f(p/q)=q$, where $p/q$ is the expression of the rational number in least terms.

Then for any $n$, the set of points at which $f$ has value less than or equal to $n$ is a subset of $bigcup_{k=1}^nfrac{1}{k}Bbb{Z}$, which is a finite union of discrete subsets of $Bbb{R}$, and hence itself a discrete subset of $Bbb{R}$.

Thus every point of $Bbb{R}$ has a punctured neighborhood disjoint from this set, and hence for every $ninBbb{N}$, and every $x_0inBbb{R}$ there exists $epsilon > 0$ such that for all $qin Bbb{Q}$ with $0<|x_0-q|<epsilon$ $f(q) > n$. However, this is exactly what it means for
$$lim_{xto x_0} f(x) = infty.$$