Limit of a rational function with a constant c

[ChiralSuperfields]

Homework Statement

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Relevant Equations

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For this problem,

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

Many thanks!

 

[anuttarasammyak]

" 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]

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

Thank you for your reply @anuttarasammyak !

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

Many thanks!

 

[anuttarasammyak]

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

 

[SammyS]

Homework Statement:: Please see below
Relevant Equations:: Please see below

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 $t\to 1$ .

 

[ChiralSuperfields]

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

Thank you for your reply @anuttarasammyak !

 

[ChiralSuperfields]

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!

 

[vela]

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]

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.

Thank you for your reply @vela!

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