$$\lim_ {x \to a^2} f(x) =L $$

Use this to prove $\lim_{t \to a} f(t^2) =L$.

This is evident from the limit substitution rule, but in this course we have not covered this and we should use the formal definition of the limit. That is:

$$ \forall \epsilon>0 \quad \exists \delta \quad s.t. \quad |x-a^2|< \delta \implies |f(x)-L|< \epsilon$$

must be used to prove:

$$ \forall \epsilon>0 \quad \exists \delta \quad s.t. \quad |t-a|< \delta \implies |f(t^2)-L|< \epsilon$$

how do I make this connection.

  • 1
    $\begingroup$ Maybe this can follow from a more general result which is not so hard to prove… (substitution theorem for limits) $\endgroup$ – user408856 Jan 28 at 9:46
  • $\begingroup$ What is that $\displaystyle\lim_{n\to\infty}$ doing in the title? $\endgroup$ – José Carlos Santos Jan 28 at 9:46

$|t-a| <\delta'$ implies $|t^{2}-a^{2}|=|t-a||t+a| <\delta' (|t|+|a|)<\delta' (\delta'+2|a|)$. So choose $\delta'$ such that $\delta' (\delta'+2|a|) <\delta$. It is enough to take $\delta' <1$ and $\delta' <\frac {\delta} {1+2|a|}$. Then $|t-a| <\delta'$ implies $|f(t^{2})-L| <\epsilon$.

  • $\begingroup$ note to self: set $x=t^2$ and it follows. $\endgroup$ – Wesley Strik Jan 28 at 10:04

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