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I'm completely lost here.

Attempt

We know that $a_n$ converges to $L$. So by the definition convergence, $\forall \varepsilon \in R, \exists M \in N$ such that $|a_n - L| < \varepsilon, \forall n \ge M$.

We need to use this fact to show that $\forall \varepsilon \in R, \exists M' \in N$ such that $|2^{a_n} - 2^L| < \varepsilon, \forall n \ge M'$

I'm completely lost as to where to start. I can't think of any properties of exponents that I can exploit here. Would appreciate some hints.

Thank You

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  • $\begingroup$ You may not get a satisfactory answer here. This is because defining exponentials (and logarithms) is a non-trivial task. It's something we are used to assuming the existence of, but not properly defining. We can say, intuitively, what $a^q$ should be for $a > 0$ and $q \in \mathbb{Q}$, but what about $a^\pi$? What does that equal? Could you provide a method to compute it? If not, then it's it's going to be hard to give a complete, rigorous, non-circular answer. $\endgroup$ – Theo Bendit Oct 22 '18 at 2:28
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We want \begin{align} |2^{a_n}-2^L| &= 2^L |2^{a_n - L} - 1|< \epsilon \end{align}

$$|2^{a_n-L}-1| < \frac{\epsilon}{2^L}$$

$$1-\frac{\epsilon}{2^L}<2^{a_n -L}< 1+\frac{\epsilon}{2^L}$$

Focus on small $\epsilon$ to make sure the lower bound is positive

$$\log _2\left(1-\frac{\epsilon}{2^L}\right)<{a_n -L}< \log_2\left( 1+\frac{\epsilon}{2^L}\right)$$

Hence we want to choose $n$ large enough that

$$|a_n-L| < \min\left(-\log _2\left(1-\frac{\epsilon}{2^L}\right), \log_2\left( 1+\frac{\epsilon}{2^L} \right)\right) $$

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Hint: $2^x$ is a continuous function (can you show this?). For a continuous function $f$ we have $f(x_n) \to f(x)$ as $x_n \to x$.

Another way: If you wanted to do $\varepsilon-\delta$ language then you could also consider $|2^{a_n}-2^L|=2^L|2^{a_n-L}-1|$

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  • $\begingroup$ If you wanted to do $\varepsilon-\delta$ language then you could also consider $|2^{a_n}-2^L|=2^L|2^{a_n-L}-1|$ $\endgroup$ – Aaron Zolotor Oct 22 '18 at 2:17
  • $\begingroup$ Thank you for the hint. This is a study question for a midterm, and we haven't talked about functions yet. So I'm guessing the professor expects us to use other means. $\endgroup$ – The_Questioner Oct 22 '18 at 2:17
  • $\begingroup$ See the edit. Let me know if you need further clarification. $\endgroup$ – Aaron Zolotor Oct 22 '18 at 2:19
  • $\begingroup$ Wouldn't it be $|1/{2^L}||2^{a_n - L} - 1|$? $\endgroup$ – The_Questioner Oct 22 '18 at 2:19
  • $\begingroup$ $2^L|2^{a_n-L}-1=2^L|\frac{2^{a_n}}{2^L}-1|$ $\endgroup$ – Aaron Zolotor Oct 22 '18 at 2:20
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A minor point of clarification in your post: just to point out that it is not the case that for any ∀ε∈R,∃M∈N such that the conditions hold. This is because ε must be positive.

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$$\lim 2^{a_n} = 2^{\lim a_n} = 2^L$$

I suggest to not use $\varepsilon-N$/$\varepsilon-\delta$ here because you would be reinventing the wheel: Instead use $\varepsilon-N$/$\varepsilon-\delta$ to prove that for a continuous function $f$, $$\lim f(a_n) = f(\lim a_n).$$

Otherwise, you'd use $\varepsilon-N$/$\varepsilon-\delta$ every time?

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