Undergrad What is the proof for nested limits?

[swampwiz]

AIUI, this is a law of proofs:

lim _x→a f( g( x ) ) = f( lim _x→a g( x ) )

I have searched for an explanation of this proof, but have been unable to find one, although I did find a page that was for certain types of functions of f( x ), just not a proof for a function in general.

 

[lurflurf]

of interest
https://teachingcalculus.com/2019/08/26/limit-of-composite-functions/

It is not true in general. That theorem is in any calculus book requiring lim x→a g( x ) exist and f continuous at that value.
In many examples of interest only one of the two conditions hold. We then need to find another method in those cases.

 

[member 587159]

Not true in general. You need continuity assumptions. Even if all the limit exists, then still it is possible to have inequality, as playing with simple examples should indicate you.

 

[swampwiz]

of interest
https://teachingcalculus.com/2019/08/26/limit-of-composite-functions/

It is not true in general. That theorem is in any calculus book requiring lim x→a g( x ) exist and f continuous at that value.
In many examples of interest only one of the two conditions hold. We then need to find another method in those cases.

I was referring to functions that are the same expression, not some contrived function that is defined by different expressions for different sections of the domain.

 

[S.G. Janssens]

I was referring to functions that are the same expression, not some contrived function that is defined by different expressions for different sections of the domain.

That is a rather arbitrary criterium. Take any function $F$ on $\mathbb{R}$ that is, in your opinion, contrived. Take another function $G$ on $\mathbb{R}$ that you don't find contrived. Define the function $H$ on $\mathbb{R}$ by $H(x) = F(G(x))$. Then $H$ can be written using one and the same expression for all $x \in \mathbb{R}$. Do you find $H$ to be contrived, or not?

 

[zinq]

Suppose that g(x) is continuous near x = a and f(u) is continuous near u = g(a). Then by definition the limit of f(g(x)) as x → a is what happens to f(g(x)) when x gets closer and closer to a. This means we are plugging into g(x) values of x close to a. By continuity of g(x) near x = a, these g(x)'s will get closer and closer to g(a). So whatever g(a) may be, the limit of f(g(x)) as x → a is the same as the limit of f(u) as u → g(a). (Are you with me so far?)

Now since f(u) is continuous at u = g(a) the limit of f(u) as u → g(a) is the same as plugging in g(a) for u, or in other words f(g(a)).