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I am trying to find the limit of the expression, the value $x$ becomes and $t$ approaches infinity;

$$x=\frac{20e^{\frac{t}{10}}-20}{2e^{\frac{t}{10}}-1}$$

I understand that the correct way to determine this limit is as follows;

$$x=\frac{20(e^{\frac{t}{10}}-1)}{2(e^{\frac{t}{10}}-\frac{1}{2})}$$ As $t$ approaches infinity, its intuitive that

$$\frac{(e^{\frac{t}{10}}-1)}{(e^{\frac{t}{10}}-\frac{1}{2})}\rightarrow 1$$ So $x$ approaches $\frac{20}{2}=1$

However, the nature of infinity is confusing me. Consider the original expression, as $t$ approaches infinity the top and bottom both approach infinity, so why can't we say the limit is one? It doesn't make sense why we can only determine the limit after doing some factorising. How does this limit process actually work?

I am assuming that the answer to this question might have to do with various sizes of infinity. It seems;

$$\frac{(e^{\frac{t}{10}}-1)}{(e^{\frac{t}{10}}-\frac{1}{2})}\rightarrow 1$$ So $x$ approaches $\frac{20}{2}=1$

However $$\frac{(e^{\frac{t}{10}}-1)}{2(e^{\frac{t}{10}}-\frac{1}{2})}\rightarrow 1$$ So $x$ approaches $\frac{20}{2}=\frac{1}{2}$

Can someone please explain this?

Quin Gardiner Bax
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    You should stop reading $x\rightarrow \infty$ as "$x$ approaches infinity" - that's not what the notation means, and it can lead you down some shady logic (like what you're asking about here). It's tricky because $x\rightarrow a$ does mean "$x$ approaches $a$," but when you write $x\rightarrow \infty$,it means that you're asking what your expression does as $x$ increases without bound. – Amaan M Sep 24 '24 at 01:47
  • Why are you claiming $\frac{20}{2}=1$ or $ \frac12$? are they typos? – user Sep 24 '24 at 09:09
  • You may want to look at this question, especially the non-accepted answer's examples of how to simplify the expression first to avoid false results - the rate of growth matters. – Dávid Laczkó Sep 24 '24 at 12:21

2 Answers2

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$$\lim_{t\to\infty} \frac {20e^\frac {t}{10} - 20}{2e^\frac {t}{10} + 2}$$

One way to resolve this is to say when $t$ is large, the size of $e^{\frac t{10}}$ dwarfs the constant terms, and the constant terms can be ignored.

$$\lim_{t\to\infty} \frac {20e^\frac {t}{10} - 20}{2e^\frac {t}{10} + 2} = \lim_{t\to\infty} \frac {20e^\frac {t}{10}}{2e^\frac {t}{10}}$$

Cancel the identical factor in the numerator and denominator.

$$\lim_{t\to\infty} \frac {20e^\frac {t}{10}}{2e^\frac {t}{10}} = \frac {20}{2} = 10$$

Some feel that this is a little too fuzzy or intuitive, and their intuition has led them astray in the past and they don't trust it.

This is a little bit more rigorous:

$$\lim_{t\to\infty} \frac {20e^\frac {t}{10} - 20}{2e^\frac {t}{10} + 2}$$

Multiply numerator and denominator by $e^{-\frac t{10}}$

$$\lim_{t\to\infty} \frac {20 - 20e^{-\frac t{10}}}{2 + 2e^\frac {-t}{10}}$$

And as $t$ goes to infinity the $e^{-\frac {t}{10}}$ terms go to $0.$

As for your question.

If we plug infinity into the numerator and denominator, why can we say that this is $\frac {\infty}{\infty} = 1$.

Because $\frac {\infty}{\infty} \ne 1!$ Well it might, but it might not. It is indeterminate, and we must work to resolve this indeterminacy.

ultralegend5385
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user317176
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Maybe, to avoid confusion and make it more clear, we can set $y=e^{\frac{t}{10}} \to \infty$ such that

$$\lim_{t\to \infty}\frac{20e^{\frac{t}{10}}-20}{2e^{\frac{t}{10}}-1}=\lim_{y\to \infty}\frac{20y-20}{2y-1}$$

this step is not necessary at all but maybe it can help to highlight what really matters for the limit, that is the "dominating" $y$ term.

From the latter expression we can conclude in other several ways, for example

  • factoring out (or dividing by) the $y$ term (the simpler)

$$\frac{20y-20}{2y-1}=\frac y y\frac{20-\frac{20}y}{2-\frac 1y}\to 10$$

  • algebraic manipulation (more tricky)

$$\frac{20y-20}{2y-1}=\frac{20y-10-10}{2y-1}=10-\frac{10}{2y-1}\to 10$$

  • squeeze theorem, eventually (overkilling)

$$\frac{20y-\sqrt y}{2y}=10-\frac1{2\sqrt y}\le \frac{20y-20}{2y-1}\le \frac{20y-10}{2y-1}=10$$

user
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