Analysis I MT22, Algebra of limits

Created: January 03, 2023 | About these notes | View in context | Study these flashcards

Flashcards

What are the two ingredients to proving that if $a _ n \to a$ and $b _ n \to b$ then $\frac{a _ n}{b _ n} \to \frac{a}{b}$?
  • Proving the product law for limits, $a _ n b _ n \to a b$
  • Proving the reciprocal law for limits, $\frac{1}{b _ n} \to b _ n$.

When proving the reciprocal law for limits, i.e. that $\frac{1}{b _ n} \to \frac{1}{b}$, what two values of epsilon do you take, to show that $ \vert b _ n - b \vert $ can be less than?
  • $\frac{ \vert b \vert }{2}$
  • $\frac{ \vert b \vert ^2}{2}\epsilon$

When proving the reciprocal law for limits, i.e. that $\frac{1}{b _ n} \to \frac{1}{b}$, what do you rearrange $\left \vert \frac{1}{b _ n} - \frac{1}{b}\right \vert $ into?
\[\begin{aligned} \left \vert \frac{1}{b _ n} - \frac{1}{b}\right \vert &= \left \vert \frac{b-b _ n}{b b _ n} \right \vert \\\\ &= \frac{1}{ \vert b \vert \vert b _ n \vert } \vert b - b _ n \vert \end{aligned}\]

When proving the reciprocal law for limits, i.e. that $\frac{1}{b _ n} \to \frac{1}{b}$, one of the aims is to show that $\frac{1}{ \vert b _ n \vert } < \frac{2}{ \vert b \vert }$. You know that $ \vert b _ n - b \vert < \frac{ \vert b \vert }{2}$, from the definition of convergence of $b _ n$. Starting from $ \vert b \vert $, how do you use this to show $\frac{1}{ \vert b _ n \vert } < \frac{2}{ \vert b \vert }$?
\[\begin{aligned} \vert b \vert &= \vert b - b _ n + b _ n \vert \\\\ &\le \vert b - b _ n \vert + \vert b _ n \vert \\\\ &< \frac{ \vert b \vert }{2} + \vert b _ n \vert \end{aligned}\]

which then implies $\frac{ \vert b \vert }{2} < \vert b _ n \vert $ and then taking reciprocals gives the required result.

When proving the product law for limits, i.e. that $a _ n b _ n \to ab$, what magic do you to $ \vert a _ n b _ n - ab \vert $ in order to get it into “a form you control”?
\[\begin{aligned} \vert a _ nb _ n - ab \vert &= \vert a _ n b _ n - ab _ n + ab _ n - ab \vert \\\\ &\le \vert a _ nb _ n - ab _ n \vert + \vert ab _ n - ab \vert \\\\ &= \vert a _ n - a \vert \vert b _ n \vert + \vert a \vert \vert b _ n - b \vert \end{aligned}\]

When proving the product law for limits, i.e. that $a _ n b _ n \to ab$, what important theorem do you make use of?

Any convergent series is bounded.



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