What is the limit of a complex fraction without using L'Hopital's rule?

[Bipolarity]

Homework Statement

Have been working 5 hours straight today. Mind is a bit fuzzy, don't know what to do here.

$$\lim_{x→1}\frac{x^{1/3}-1}{\sqrt{x}-1}$$

Homework Equations

The Attempt at a Solution

I trying multiplying by conjugates. Did not work. Also, no L'Hopital allowed.
I can't believe it because I solved problems much harder than this today, yet this problem is stopping me. Sigh. Any ideas, guys?

BiP

 

[micromass]

I trying multiplying by conjugates. Did not work.

Can you show us what you did?

 

[LCKurtz]

Hint: Let $x=u^6$.

 

[Bipolarity]

Hint: Let $x=u^6$.

Ah I see. But why are we allowed to change variables in limits?

BiP

 

[LCKurtz]

Hint: Let $x=u^6$.

Ah I see. But why are we allowed to change variables in limits?

BiP

Because I give you permission :uhh:

Seriously, that is a good question. It depends on what theorems you have or are allowed to use. Most first year calculus texts proceed pretty informally with that subject. If you want to see some of the technical details, one place to look is here:

http://www.swarthmore.edu/NatSci/smaurer1/Math18H/inv_lim.pdf

but it might be more than you want to know, depending on your level.

 

[HallsofIvy]

Hint: Let $x=u^6$.

Ah I see. But why are we allowed to change variables in limits?

BiP

You can, of course, always replace a term with something that is equal to it. That's a basic concept of arithmetic as when we replace "1+ 1" with "2".

I suppose you see that LCKurtz's suggested substitution changes your problem to $(u^2- 1)/(u^3- 1)$ and you can factor u- 1 out of each term.

You could do the same thing without actually substituting. You know that $u^3- 1= (u- 1)(u^2+ u+ 1)$ which is the same as saying that $x- 1= (x^{1/3}- 1)(x^{2/3}+ x^{1/3}+ 1)$ and that $u^2- 1= (u- 1)(u+ 1)$ which is the same as saying that $x-1= (x^{1/2}- 1)(x^{1/2}+ 1)$.

So you can multiply numerator and denominator of $(x^{1/3}- 1)/(x^{1/2}- 1)$ by both $x^{2/3}+ x^{1/3}+ 1$ and $x^{1/2}+ 1$ to get
$$\frac{x- 1}{x- 1}\frac{x^{1/2}+ 1}{x^{2/3}+ x^{1/2}+ 1}$$
and, of course, cancel the "x- 1" terms before taking the limit.

 

[Bipolarity]

You can, of course, always replace a term with something that is equal to it. That's a basic concept of arithmetic as when we replace "1+ 1" with "2".

I suppose you see that LCKurtz's suggested substitution changes your problem to $(u^2- 1)/(u^3- 1)$ and you can factor u- 1 out of each term.

You could do the same thing without actually substituting. You know that $u^3- 1= (u- 1)(u^2+ u+ 1)$ which is the same as saying that $x- 1= (x^{1/3}- 1)(x^{2/3}+ x^{1/3}+ 1)$ and that $u^2- 1= (u- 1)(u+ 1)$ which is the same as saying that $x-1= (x^{1/2}- 1)(x^{1/2}+ 1)$.

So you can multiply numerator and denominator of $(x^{1/3}- 1)/(x^{1/2}- 1)$ by both $x^{2/3}+ x^{1/3}+ 1$ and $x^{1/2}+ 1$ to get
$$\frac{x- 1}{x- 1}\frac{x^{1/2}+ 1}{x^{2/3}+ x^{1/2}+ 1}$$
and, of course, cancel the "x- 1" terms before taking the limit.

Thank you Ivy! Indeed, I have always believed that any probem that can be solved with a substitution can also be solved without the substitution, since the substitution serves merely to simplify the nomenclature or intuition of the problem.

But it still seems relevant that the change of variables in limits is a dubious issue. I wonder why it is not usually addressed rigorously in standard calculus/analysis texts?

BiP