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Are the following expressions equivalent?: $\lim_{3x \rightarrow 5}x^2+1$ and $\lim_{x \rightarrow 5/3}x^2+1$

JJL
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  • yes, they are equivalent. – user474330 Oct 05 '17 at 03:09
  • Yes. You can mimic what I did here to prove it formally. – Ivo Terek Oct 05 '17 at 03:10
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    Some people will say that the first is improperly written, that one should only have a limiting variable appear without coefficients or other modifications. If you were to relax this and allow for your limiting variable to be more than "just $x$" then yes it is fine. I would still avoid ever writing the one on the left though. – JMoravitz Oct 05 '17 at 03:27
  • The first expression is not standard. At least I have never seen it written in well known books. If you have seen this notation, it would be nice to see a definition of $\lim_{f(x) \to a} g(x) =L$. Also it needs a proper development of corresponding limit laws and most importantly what advantages does it offer in comparison to the standard limit $\lim_{x\to a} f(x) =L$. – Paramanand Singh Oct 05 '17 at 07:12

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A quick observation: $|3x-5| = \bigg|3\cdot (x-\dfrac{5}{3})\bigg | = 3\bigg|x-\dfrac{5}{3}\bigg|$

Let us prove that if $$\lim_{3x\to 5} x^2+1 = \dfrac{34}{9}$$ then $$\lim_{x\to \frac{5}{3}} x^2+1 = \dfrac{34}{9}$$

$\lim_{3x\to 5} x^2+1 = \dfrac{34}{9}$ implies that,

For a given $\epsilon >0$, $\exists \delta>0$ such that $\forall x$ with $0<|3x-5|<\delta$, $|x^2+1-\dfrac{34}{9}|<\epsilon$

As mentioned in the first line, $|3x-5| = 3\bigg|x-\dfrac{5}{3}\bigg|$.

So, $0<|3x-5|<\delta$ $\iff$ $0<\bigg|x-\dfrac{5}{3}\bigg|< \dfrac{\delta}{3}$.

The inequality tells us to choose $\delta'$ as $\delta'=\dfrac{\delta}{3}$.

And we can immediately see that, $0<\bigg|x-\dfrac{5}{3}\bigg|<\delta' \implies 0<|3x-5|<\delta \implies |x^2+1-\dfrac{34}{9}|<\epsilon$

Hence, by the definition of limit, $$\lim_{x\to \frac{5}{3}} x^2+1 = \dfrac{34}{9}$$

Now, you can easily prove the other direction, that is

if $$\lim_{x\to \frac{5}{3}} x^2+1 = \dfrac{34}{9}$$ then $$\lim_{3x\to 5} x^2+1 = \dfrac{34}{9}$$

which completes the proof that

$$\lim_{3x\to 5} x^2+1 = \dfrac{34}{9} \iff \lim_{x\to \frac{5}{3}} x^2+1 = \dfrac{34}{9}$$

yh016
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