Limits of functions, inequalities (analysis)

[Kate2010]

Homework Statement

Assume:
1) f(x) $$\rightarrow$$ y_o as x $$\rightarrow$$ x₀
2) g(y)$$\rightarrow$$ l as y $$\rightarrow$$ y₀
3) g(y₀) = l

Prove that g(f(x)) $$\rightarrow$$ l as x $$\rightarrow$$ x₀

Homework Equations

Definition of function tending to limit - E is a subset of R, f:E->R, f tends to l as x tends to p if for all e>0 there exists a d>0 such that |f(x) - l | < e for all x in E such that 0 < |x-p|< d.

Theorem: f: E $$\rightarrow$$ R where E $$\subseteq$$ R, p is a limit point of E and l $$\in$$ R. The following are equivalent:
a) f(x) $$\rightarrow$$ l as x $$\rightarrow$$ p
b) for every sequence {p_n} in E such that p_n $$\neq$$ p and lim$$_{n\rightarrow\infty}$$ p_n = p we have that f(p_n) $$\rightarrow$$ l as n$$\rightarrow$$ $$\infty$$

The Attempt at a Solution

I thought I could use the above theorem with (1) to say that every sequence {x_n} such that x_n$$\neq$$ x₀ and lim$$_{n\rightarrow\infty}$$ x_n = x₀, we have that f(x_n) $$\rightarrow$$ y₀ as n $$\rightarrow$$ $$\infty$$

Similarly from (2) every sequence {y_n} such that y_n$$\neq$$ y₀ and lim$$_{n\rightarrow\infty}$$ y_n = y₀, we have that g(y_n) $$\rightarrow$$ l as n $$\rightarrow$$ $$\infty$$

So, can I let y_n = f(x) ( or maybe f(x_n)?
Would this help show that g(f(x)) $$\rightarrow$$ l? I haven't used (3) or really shown this happens as x $$\rightarrow$$ x₀.

 

[ystael]

You don't need sequences to do this; you can use the definition of limit directly. Find a condition on $$y$$ that ensures that $$g(y)$$ is close to $$l$$, then find a condition on $$x$$ that ensures that $$f(x)$$ satisfies that condition on $$y$$.

 

[Kate2010]

From (3) I know for all $$\epsilon$$>0 $$\exists$$ $$\delta$$₁ > 0 such that |g(y)-l|<$$\epsilon$$ for all y such that 0<|y-y₀|<$$\delta$$₁

Similarly from (1)
for all $$\epsilon$$>0 $$\exists$$ $$\delta$$₂ > 0 such that |f(x)-y₀|<$$\epsilon$$ for all x such that 0<|x-x₀|<$$\delta$$₂

Can I then do something along these lines:

|g(f(x)-y₀)-l|
=|g(f(x)) - g(y₀) -l|
=|g(f(x)) - 2l|

Is less than $$\epsilon$$ for all $$\delta$$=min($$\delta$$₁, $$\delta$$₂) such that 0<|x-x₀|<$$\delta$$?

 

[Altabeh]

From (3) I know for all $$\epsilon$$>0 $$\exists$$ $$\delta$$₁ > 0 such that |g(y)-l|<$$\epsilon$$ for all y such that 0<|y-y₀|<$$\delta$$₁

Similarly from (1)
for all $$\epsilon$$>0 $$\exists$$ $$\delta$$₂ > 0 such that |f(x)-y₀|<$$\epsilon$$ for all x such that 0<|x-x₀|<$$\delta$$₂

Can I then do something along these lines:

|g(f(x)-y₀)-l|
=|g(f(x)) - g(y₀) -l|
=|g(f(x)) - 2l|

Is less than $$\epsilon$$ for all $$\delta$$=min($$\delta$$₁, $$\delta$$₂) such that 0<|x-x₀|<$$\delta$$?

They are not correct because one can't ensure that g(f(x)-y₀) is close to $$l$$ and of course g(y) is not a linear function to assign such a property to it that g(f(x)-y₀) = g(f(x))-g(y₀). Try to apply both the f(x) and g(f(x)) to the triangular inequality!

AB

 

[Kate2010]

How is this?

Let $$\epsilon$$> 0
$$\exists$$$$\delta$$₁>0 such that |g(y)-l| = |g(y) - g(y₀| < $$\epsilon$$
$$\exists$$ ]$$\delta$$₂>0 such that |f(x) - y₀| < $$\delta$$₂
So for all x such that |x-x₀|< min($$\delta$$₁,$$\delta$$₂) we have
|(gof)(x) - (gof)(x₀)| = |g(f(x))-g(f(x₀))| = |g(f(x)) - l| < $$\epsilon$$