Limits of functions, inequalities (analysis)

  • Thread starter Kate2010
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  • #1
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Homework Statement



Assume:
1) f(x) [tex]\rightarrow[/tex] yo as x [tex]\rightarrow[/tex] x0
2) g(y)[tex]\rightarrow[/tex] l as y [tex]\rightarrow[/tex] y0
3) g(y0) = l

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

The Attempt at a Solution



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

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

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

Answers and Replies

  • #2
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You don't need sequences to do this; you can use the definition of limit directly. Find a condition on [tex]y[/tex] that ensures that [tex]g(y)[/tex] is close to [tex]l[/tex], then find a condition on [tex]x[/tex] that ensures that [tex]f(x)[/tex] satisfies that condition on [tex]y[/tex].
 
  • #3
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From (3) I know for all [tex]\epsilon[/tex]>0 [tex]\exists[/tex] [tex]\delta[/tex]1 > 0 such that |g(y)-l|<[tex]\epsilon[/tex] for all y such that 0<|y-y0|<[tex]\delta[/tex]1

Similarly from (1)
for all [tex]\epsilon[/tex]>0 [tex]\exists[/tex] [tex]\delta[/tex]2 > 0 such that |f(x)-y0|<[tex]\epsilon[/tex] for all x such that 0<|x-x0|<[tex]\delta[/tex]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 [tex]\epsilon[/tex] for all [tex]\delta[/tex]=min([tex]\delta[/tex]1, [tex]\delta[/tex]2) such that 0<|x-x0|<[tex]\delta[/tex]?
 
  • #4
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From (3) I know for all [tex]\epsilon[/tex]>0 [tex]\exists[/tex] [tex]\delta[/tex]1 > 0 such that |g(y)-l|<[tex]\epsilon[/tex] for all y such that 0<|y-y0|<[tex]\delta[/tex]1

Similarly from (1)
for all [tex]\epsilon[/tex]>0 [tex]\exists[/tex] [tex]\delta[/tex]2 > 0 such that |f(x)-y0|<[tex]\epsilon[/tex] for all x such that 0<|x-x0|<[tex]\delta[/tex]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 [tex]\epsilon[/tex] for all [tex]\delta[/tex]=min([tex]\delta[/tex]1, [tex]\delta[/tex]2) such that 0<|x-x0|<[tex]\delta[/tex]?

They are not correct because one can't ensure that g(f(x)-y0) is close to [tex]l[/tex] 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
146
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How is this?

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

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