Proof of the limit's composition law when x approaches infinity

Theorem: If f(x) is continous at b and limxg(x)=b, then

limxf(g(x))=f(b)=f(limxg(x))

Proof: Because f(x) is continous at b, so

limxbf(x)=f(b)

Because limxg(x)=b

Combine (1) and (2), we get

limxf(g(x))=f(b)

Thus we proved the right side of equation (1)

Now we need to prove limxf(g(x))=f(b)

Because

limxbf(x)=f(b)

according to the definition of limit, ε>0, there exists a δ such that for all |xb|<δ, we have

|f(x)f(b)|<ε

Replace x with g(x) in the above conclusion, we get

ε>0, there exists a δ such that for all |g(x)b|<δ, we have

|f(g(x))f(b)|<ε

Note that although f(x) needs to be defined and has a limited value (6) around b, g(x) doesn’t need to be so. For example, g(x) may never be larger than b. But every time we get a g(x) that meets the condition |g(x)b|<δ, |f(g(x))f(b)|<ε is guaranteed.

Because

limxg(x)=b

according to the definition of limit, ε>0, there exists a δ such that for all x>δ we have |g(x)b|<ε

Let ε=δ, so x>δ, we have

|g(x)b|<δ

Combine this with (6), ε>0, there exists δ, whenever x>δ, we have

|f(g(x))f(b)|<ε

Which is exactly the definition of the limit

limxf(g(x))=f(b)

So we proved the left side of (1) equation. So equation

limxf(g(x))=f(b)=f(limxg(x))

holds.

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