# Limit of a rational function with a constant c

• ChiralSuperfields
In summary: I will take note of that.In summary, the conversation involves a discussion on the relationship between ##x## and ##t## as they approach zero and one respectively. The formula given is ##\lim_{t \rightarrow 1} \frac{c(t-1)}{t^3-1}## and it is shown that as ##x\to 0##, ##t\to 1##. However, the reasoning stated in the conversation is incorrect as it involves an incorrect distribution of the cube root and an incorrect use of the notation ##\propto##.
ChiralSuperfields
Homework Statement
Relevant Equations
For this problem,

Did they get ## x## approaches one is equivalent to ##t## approaches zero because ##t ∝ (x)^{1/3} + 1##?

Many thanks!

Last edited:
" Notice that x ##\rightarrow##0 is equivalent to t ##\rightarrow## 1 ", it says. The given formula becomes
$$\lim_{t \rightarrow 1} \frac{c(t-1)}{t^3-1}$$

ChiralSuperfields
anuttarasammyak said:
" Notice that x ##\rightarrow##0 is equivalent to t ##\rightarrow## 1 ", it says. The given formula becomes
$$\lim_{t \rightarrow 1} \frac{c(t-1)}{t^3-1}$$

Sorry, I have updated the question after I realized my mistake. Is my reasoning correct though?

Many thanks!

$$\lim_{x\rightarrow 0}t=\lim_{x\rightarrow 0} \sqrt[3]{1+cx}=1$$
$$\lim_{t\rightarrow 1}x=\lim_{t\rightarrow 1}\frac{t^3-1}{c}=0$$

ChiralSuperfields
Callumnc1 said:

For this problem,
View attachment 322282
Did they get ## x## approaches one is equivalent to ##t## approaches zero because ##t ∝ (x)^{1/3} + 1##?

Many thanks!
I would say it's more like:

As ##x\to 0##, it's clear that ##\displaystyle \root 3 \of{1+cx \,} \to 1##, so ##t\to 1## .

Last edited:
ChiralSuperfields
anuttarasammyak said:
$$\lim_{x\rightarrow 0}t=\lim_{x\rightarrow 0} \sqrt[3]{1+cx}=1$$
$$\lim_{t\rightarrow 1}x=\lim_{t\rightarrow 1}\frac{t^3-1}{c}=0$$

SammyS said:
I would say it's more like:

As ##x\to 0##, it's clear that ##\root 3 \of{1+cx \,} \to 1##, so ##t\to 1## .
Thank you @SammyS , I see now!

Callumnc1 said:
Is my reasoning correct though?
It looks like you're thinking ##\sqrt[3]{1+cx} = \sqrt[3]{1} + \sqrt[3]{cx}##. That's clearly wrong. You can't distribute the root across the addition.

Also, as far as notation goes, ##\propto## means "proportional to", so saying that ##t \propto 1 + x^{1/3}## means that ##t = k(1+x^{1/3})## for some constant ##k##, which you probably didn't mean.

ChiralSuperfields and PeroK
vela said:
It looks like you're thinking ##\sqrt[3]{1+cx} = \sqrt[3]{1} + \sqrt[3]{cx}##. That's clearly wrong. You can't distribute the root across the addition.

Also, as far as notation goes, ##\propto## means "proportional to", so saying that ##t \propto 1 + x^{1/3}## means that ##t = k(1+x^{1/3})## for some constant ##k##, which you probably didn't mean.

That is good you mentioned the notation, I didn't realize I could not do that!

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