Can You Prove Lim(x->0) g(x) Equals L When g(x)=f(ax)?

[rohitmishra]

Suppose f:R->R , Lim (x->0) f(x) = L. and a>0.

Define g:R->R by g(x)= f(ax). Prove that Lim (x->0) g(x) = L ?


My Solution.

I am not able to prove it as equal to L.

From what I get it is

Lim (x->0) g(x)

Lim (x->0) a * f(x)

a* Lim (x->0) f(x)

a* L = aL ??

I have to prove Lim (x->0) g(x) = L.

Help me out

 

[slider142]

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Define g:R->R by g(x)= f(ax). Prove that Lim (x->0) g(x) = L ?
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Lim (x->0) g(x)

Lim (x->0) a * f(x)

I do not see how you get the equality f(ax) = a*f(x). This is only true when f is linear on R.

If you do not have any theorems dealing with limits of compositions, then it is still easy to see that ax approaches 0 whenever x approaches 0. You have to rigorize this statement by noting that g(x) = f(u) when u = ax. The fact that the limit exists for f(u) as u approaches 0 gives you a predetermined delta neighborhood of u for any epsilon. You have to then show what delta neighborhood you should use for x = u/a.