# Improper integral defined as one that the integral have one or more discontinuities or infinity.

Here is the function $f(x)=\frac{1}{x}$.

f(x) have the limit as x approach to infinity (0) but f(x) integral from 0 to inf is infinity : $\int _0^{\infty }\:\frac{1}{x}dx\:=\:\infty$

If a integral do have a limit, and the integral is an improper integral. so the integral from 0 to inf must convergent.

You're confusing the limit of $f(x)$ and the limit of the integral over $[1,x]$.
The integral $\displaystyle{\int_1^{\infty} f(x)\,dx}$ is said to be convergent when $$\lim_{X\rightarrow\infty}\int_1^{\infty} f(x)\,dx$$ exists, which is not the same as whether $f(x)$ itself tends to $0$ as $x$ tends to infinity.
You can actually construct functions such that $\displaystyle{\int_1^{\infty} f(x)\,dx}$ is convergent but $f(x)$ has no limit as $x\rightarrow\infty$.