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 \delta₁ > 0 such that |g(y)-l|<\epsilon for all y such that 0<|y-y₀|<\delta₁

Similarly from (1)
for all \epsilon>0 \exists \delta₂ > 0 such that |f(x)-y₀|<\epsilon for all x such that 0<|x-x₀|<\delta₂

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(\delta₁, \delta₂) such that 0<|x-x₀|<\delta?

 

[Altabeh]

From (3) I know for all \epsilon>0 \exists \delta₁ > 0 such that |g(y)-l|<\epsilon for all y such that 0<|y-y₀|<\delta₁

Similarly from (1)
for all \epsilon>0 \exists \delta₂ > 0 such that |f(x)-y₀|<\epsilon for all x such that 0<|x-x₀|<\delta₂

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(\delta₁, \delta₂) 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\delta₁>0 such that |g(y)-l| = |g(y) - g(y₀| < \epsilon
\exists ]\delta₂>0 such that |f(x) - y₀| < \delta₂
So for all x such that |x-x₀|< min(\delta₁,\delta₂) we have
|(gof)(x) - (gof)(x₀)| = |g(f(x))-g(f(x₀))| = |g(f(x)) - l| < \epsilon