# Limits of functions, inequalities (analysis)

1. Jan 22, 2010

### Kate2010

1. The problem statement, all variables and given/known data

Assume:
1) f(x) $$\rightarrow$$ yo 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

2. Relevant 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 {pn} in E such that pn $$\neq$$ p and lim$$_{n\rightarrow\infty}$$ pn = p we have that f(pn) $$\rightarrow$$ l as n$$\rightarrow$$ $$\infty$$

3. The attempt at a solution

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

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

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

2. Jan 22, 2010

### 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 $$y$$.

3. Jan 23, 2010

### 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|<$$\delta$$?

4. Jan 23, 2010

### Altabeh

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

5. Jan 23, 2010

### 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| < $$\epsilon$$