First, please read this summary of what we do, how we do it, and what you must do to get help.
https://www.freemathhelp.com/forum/t...l=1#post433156
In this case, we particularly need to know whether you have studied limits in a calculus or a pre-calculus course.
Second, we need more context because what you have shown us is odd. Perhaps there is more that you have not shown.
Third, please tell us what you have tried or thought while trying to understand.
EDIT: What you have been shown is not math, but intuition.
If x is very large, then
[tex]\dfrac{2(x + a + 1) - b}{x} = \dfrac{2x}{x} + \dfrac{2a + 2 - b}{x} \approx 2 + 0 = 2.[/tex]
Similarly, if x is very large,
[tex]\dfrac{x + 2a + 2}{x} = \dfrac{x}{x} + \dfrac{2a + 2}{x} \approx 1 + 0 = 1.[/tex]
So if x is very large
[tex]\dfrac{2(x + a + 1) - b}{(x + 1)(x + 2a + 2} = \dfrac{\dfrac{2(x + a + 1) - b}{x}}{ (x + 1) * \dfrac{x + 2a + 2}{x}} \approx \dfrac{2}{(x + 1) * 1} = \dfrac{2}{x + 1}.[/tex]
This is kind of foolish however because if x is very large, the resulting approximation is itself is approximately zero.
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