another interesting limit

[Darya]

I'm having trouble with the following:
[MATH] \lim_{x \to 1} (x-2x^{1/100}+1)/(x^{1/2}-2x^{1/100}+1) [/MATH]Please help with a hint. I tried substitution, rewriting it as (x-x^1/100)+(1--x^1/100) and so on in order to use A^2-B^2 formula but nothing works...

So sorry! Edit: it is not allowed to use LHopitals rule

 

[Deleted member 4993]

I'm having trouble with the following:
[MATH] \lim_{x \to 1} (x-2x^{1/100}+1)/(x^{1/2}-2x^{1/100}+1) [/MATH]Please help with a hint. I tried substitution, rewriting it as (x-x^1/100)+(1--x^1/100) and so on in order to use A^2-B^2 formula but nothing works...

Have you tried L'Hopital's method?

 

[HallsofIvy]

Both numerator and denominator are 0 at x= 1 so L'Hopital's rule applies.

 

[Darya]

Have you tried L'Hospital's method?

edit: it's not allowed

 

[Darya]

Both numerator and denominator are 0 at x= 1 so L'Hopital's rule applies.

edit: it's not allowed

 

[Steven G]

You want to multiply the denominator (and numerator!) by something so that the limit as x->1 of the denominator is not zero. Think conjugate.

 

[Romsek]

If Jomo's method in post #6 works, use it. I couldn't find a multiplier that accomplishes what he suggests.

Two things that make things a bit easier I think.

i) [MATH]u = x^{1/100} \Rightarrow \dfrac{u^{100}-2u+1}{u^{50}-2u+1}[/MATH]
[MATH]x \to 1 \Rightarrow u \to 1[/MATH]
ii) [MATH]\dfrac{u^{100}-2u+1}{u^{50}-2u+1} = (u^{50}+2 u-1) + \dfrac{4 u^2-6 u+2}{u^{50} - 2 u + 1}[/MATH]
[MATH]\lim \limits_{u\to 1} \left(u^{50}+2 u-1 + \dfrac{4 u^2-6 u+2}{u^{50} - 2 u + 1}\right) = 2 + \lim \limits_{u\to1} \dfrac{4 u^2-6 u+2}{u^{50} - 2 u + 1}[/MATH]
You still have a limit to find by some means but it's a bit less unwieldy.

 

[Romsek]

If Jomo's method in post #6 works, use it. I couldn't find a multiplier that accomplishes what he suggests.

Two things that make things a bit easier I think.

i) [MATH]u = x^{1/100} \Rightarrow \dfrac{u^{100}-2u+1}{u^{50}-2u+1}[/MATH]
[MATH]x \to 1 \Rightarrow u \to 1[/MATH]
ii) [MATH]\dfrac{u^{100}-2u+1}{u^{50}-2u+1} = (u^{50}+2 u-1) + \dfrac{4 u^2-6 u+2}{u^{50} - 2 u + 1}[/MATH]
[MATH]\lim \limits_{u\to 1} \left(u^{50}+2 u-1 + \dfrac{4 u^2-6 u+2}{u^{50} - 2 u + 1}\right) = 2 + \lim \limits_{u\to1} \dfrac{4 u^2-6 u+2}{u^{50} - 2 u + 1}[/MATH]
You still have a limit to find by some means but it's a bit less unwieldy.

Ok the next thing I'd do is a nasty little trick.

You want to basically expand the numerator and denominator as Taylor series about [MATH]u-1[/MATH]We only need the linear terms so we don't really need to use derivatives... we can pretend we're just dividing alot..

[MATH]4u^2-6u+2 = 2(u-1) + 4(u-1)^2 \\ u^{50}-2u + 1 = 48(u-1) + 1225(u-1)^2 + \dots[/MATH]
[MATH]\dfrac{4 u^2-6 u+2}{u^{50} - 2 u + 1} = \dfrac{2(u-1) + 4(u-1)^2}{48(u-1) + 1225(u-1)^2 + \dots}[/MATH]
Now we pull our usual "divide by the highest power of [MATH]x[/MATH]" trick, but here [MATH]x=u-1[/MATH]
we end up with [MATH]\lim \limits_{u\to1} \dfrac{4 u^2-6 u+2}{u^{50} - 2 u + 1} = \dfrac{2}{48}=\dfrac{1}{24}[/MATH]
then adding the 2 from the previous section we have

[MATH]\lim \limits_{u\to 1} \dfrac{u^{100}-2u+1}{u^{50}-2u+1} = 2 + \dfrac{1}{24} = \dfrac{49}{24}[/MATH]
Now of course I have waved a wand to produce the 48 that appears above. You have to figure out how to obtain that.

 

[Deleted member 4993]

You want to multiply the denominator (and numerator!) by something so that the limit as x->1 of the denominator is not zero. Think conjugate.

Although I liked your idea, but I (like Romsek) could not find a useful conjugate.

 

[Darya]

Although I liked your idea, but I (like Romsek) could not find a useful conjugate.

i also sought a suitable conjugate at first. Since Taylor series (although I'm familiar with them) have not been in the course yet, it's not allowed to use them either :0 I'll text if I come up with something Have a nice day, thanks for help!

 

[lookagain]

Using post # 7, I'm going to show some steps in factoring the algebraic fraction,
that the limit is being taken of, as an alternate method.

\(\displaystyle \dfrac{2(2u^2 - 3u + 1)}{u^{50} - u^{49} + u^{49} - u^{48} + u^{48} \ - \ ... \ - \ u^3 + u^3 - u^2 + u^2 - 2u + 1} \ =\)

\(\displaystyle \dfrac{2(u - 1)(2u - 1)}{u^{49}(u -1) \ + \ u^{48}(u - 1) \ + \ ... \ + \ u^2(u - 1) \ + \ (u - 1)(u - 1)} \ =\)

\(\displaystyle \dfrac{2(u - 1)(2u - 1)}{(u - 1)(u^{49} + u^{48} \ + \ ... \ + \ u^2 + u^1 - 1)} \ = \ ?\)

In front of these fractions is the limit as u approaches 1. And this partial result is
added to 2 for the full answer.

Darya, can you complete the end of the limit steps of my alternative to arrive at Romsek's answer?

 

[Steven G]

How did we all miss factoring out u-1. That was routine!

 

[Darya]

Using post # 7, I'm going to show some steps in factoring the algebraic fraction,
that the limit is being taken of, as an alternate method.

\(\displaystyle \dfrac{2(2u^2 - 3u + 1)}{u^{50} - u^{49} + u^{49} - u^{48} + u^{48} \ - \ ... \ - \ u^3 + u^3 - u^2 + u^2 - 2u + 1} \ =\)

\(\displaystyle \dfrac{2(u - 1)(2u - 1)}{u^{49}(u -1) \ + \ u^{48}(u - 1) \ + \ ... \ + \ u^2(u - 1) \ + \ (u - 1)(u - 1)} \ =\)

\(\displaystyle \dfrac{2(u - 1)(2u - 1)}{(u - 1)(u^{49} + u^{48} \ + \ ... \ + \ u^2 + u^1 - 1)} \ = \ ?\)

In front of these fractions is the limit as u approaches 1. And this partial result is
added to 2 for the full answer.

Darya, can you complete the end of the limit steps of my alternative to arrive at Romsek's answer?

Yes! Alternatively, with substitution from post 7 and then long division by (u-1) we get (u^99+u^98+u^97+...+u-1)/((u^49+u^48+u^47+...+u-1), which is equal to 98/48 or 49/24. Thaaaankssss!!!!

 

[Romsek]

The key point to all this is that we converted the problem from one where the limit ended up with a bunch of terms that did not go to zero in the limit to a problem where most of the terms did go to zero in the limit. Thus we were sort of able to pluck off the important terms.

Taking a step back this is all basically why L'Hopital's rule works.