You don't. Not really. The integral does not exist as integral, you have to apply it as a distribution to a suitable test function.
What the formula then really means is the Fourier inversion theorem, i.e., that the inverse Fourier transform of the Fourier transform of a suitably smooth and bounded function returns the same function. This is a very deep result in Fourier analysis, you can not hope to solve it with a simple limit.
If you want an intuitive way to understand it, approximate the constant 1 by a (slowly) exponentially decaying function, i.e., consider for small $0<α$
$$\int_{-\infty}^{\infty} e^{-α|b|}\cdot e^{ibx}\,db= \frac{2α}{α^2+x^2}$$
which gives a proper approximation of the delta distribution, positive function values, constant integral over $\mathbb R$ with value $2\pi$ and pointwise convergence to zero except in $x=0$.
This approximation is usually also used in the proof of the Fourier inversion theorem.
Another familiar approximation uses the Gauss bell function, again for small positive $α$
$$\int_{-\infty}^{\infty} e^{-\tfracα2 b^2}\cdot e^{ibx}\,db
=\int_{-\infty}^{\infty} e^{-\tfracα2 (b-i\tfrac{x}{α})^2}\cdot e^{-\tfrac{x^2}{2α}}\,db=\sqrt{\tfrac{2\pi}{α}}e^{-\tfrac{x^2}{2α}}$$
gives the same properties of an delta approximating family, more exactly, it converges as family of tempered distributions to $2\pi\delta_0$.
For a nice presentation of the necessary details see Carl Offner: "A little harmonic analysis"
