Theorem: If f(x) is continous at b and limx→∞g(x)=b, then
limx→∞f(g(x))=f(b)=f(limx→∞g(x))
Proof: Because f(x) is continous at b, so
limx→bf(x)=f(b)
Because limx→∞g(x)=b
Combine (1) and (2), we get
limx→∞f(g(x))=f(b)
Thus we proved the right side of equation (1)
Now we need to prove limx→∞f(g(x))=f(b)
Because
limx→bf(x)=f(b)
according to the definition of limit, ∀ε>0, there exists a δ′ such that for all |x−b|<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
limx→∞g(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
limx→∞f(g(x))=f(b)
So we proved the left side of (1) equation. So equation
limx→∞f(g(x))=f(b)=f(limx→∞g(x))
holds.