Solving Indeterminate Forms: Evaluating Limits with Exponential Functions

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (3^y - 5^y) / (2^y - 7^y) as y approaches 0, which is identified as an indeterminate form [0/0]. Participants express frustration over the complexity of the problem and share their attempts to manipulate the expression to resolve the indeterminacy.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss factoring terms out of the numerator and denominator, exploring the use of logarithmic properties, and considering the application of special limits. There are questions about how to eliminate the variable y from the expression and concerns about the validity of their approaches.

Discussion Status

The conversation is ongoing, with participants sharing various strategies and expressing confusion over their results. Some have suggested using specific limits, while others have noted their struggles with logarithmic manipulations. There is a mix of progress and setbacks, with no clear consensus on the best approach yet.

Contextual Notes

Participants mention time constraints due to upcoming exams, which may be affecting their ability to work through the problem thoroughly. The repeated mention of the indeterminate form indicates a shared understanding of the challenge at hand.

alingy1
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lim of (3^y-5^y)/(2^y-7^y) as y->0

Evaluate the above limit.

Okay, so, I was flabbergasted at how challenging this problem is. I realized it was an indeterminate form [0/0]. I was trying to find a way to cancel y out of the numerator and denominator, in vain. I tried finding a pattern to expand the numerator and denominator, in vain: I still ended up with the [0/0].

Any tips?

Please, I have been racking my brains for 3.5 hours. :(
 
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alingy1 said:
lim of (3^y-5^y)/(2^y-7^y) as y->0

Evaluate the above limit.

Okay, so, I was flabbergasted at how challenging this problem is. I realized it was an indeterminate form [0/0]. I was trying to find a way to cancel y out of the numerator and denominator, in vain. I tried finding a pattern to expand the numerator and denominator, in vain: I still ended up with the [0/0].

Any tips?

Please, I have been racking my brains for 3.5 hours. :(

I would start by factoring a ##3^y## out of the numerator and a ##2^y## out of the denominator giving$$
\left (\frac 3 2\right)^y \frac{1-(\frac 5 3)^y}{1 -(\frac 7 2)^y}$$and see if that gives you any ideas.
 
:S
I feel really lost and desperate :S.
I tried taking the (5/3)^y and the (7/2)^y out again. No results. I'm trying to find a way of getting rid of that y.

I have the answer in front of my face: ln5-ln3 over ln7-ln2
I don't see how they get those logs in there :S

Please be patient with my slow mind :P I have been studying from 8AM to 9PM.

EDIT: I BELIEVE I AM GETTING SOMEWHERE. :)
 
Last edited:
alingy1 said:
:S
I feel really lost and desperate :S.
I tried taking the (5/3)^y and the (7/2)^y out again. No results. I'm trying to find a way of getting rid of that y.

I have the answer in front of my face: ln5-ln3 over ln7-ln2
I don't see how they get those logs in there :S

Please be patient with my slow mind :P I have been studying from 8AM to 9PM.

EDIT: I BELIEVE I AM GETTING SOMEWHERE. :)

Use your 'special limit'. Have you figured out what lim x->0 (a^x-1)/x is yet? It's 1 if a=e. What is it in general? That's a great place to start.
 
Okay, I am lost.

I made a mistake using the logarithm rules. I came to an answer that was close to the right answer, which excited me a bit. :(

Dick: I just don't see how I could get y at the denominator to use that special limit.
 
alingy1 said:
Okay, I am lost.

I made a mistake using the logarithm rules. I came to an answer that was close to the right answer, which excited me a bit. :(

Dick: I just don't see how I could get y at the denominator to use that special limit.

Divide the numerator AND the denominator of LCKurtz's fraction by y. It doesn't change the result, right?
 
Yes, I could find the answer to my previous question. Turns out I didn't need that 'special limit'. Indeed, I just multiplied the denominator and numerator by a^x and that did the trick in me elucidating that daunting question. Sorry I didn't have time to follow up! I have exams this week, so it's rush week!
 
alingy1 said:
Yes, I could find the answer to my previous question. Turns out I didn't need that 'special limit'. Indeed, I just multiplied the denominator and numerator by a^x and that did the trick in me elucidating that daunting question. Sorry I didn't have time to follow up! I have exams this week, so it's rush week!

It's fine if you don't have time to follow up. But what did you use instead of the 'special limit'? I don't think just multiplying by a^x will do it. If you think you've got a solution that way it's likely wrong. Just saying...
 
Could you use L'Hopitals?
 
  • #10
Oh no, do not worry. I got the right answer. Actually, I may have used the 'special limit'. I don't recall. I will have to search through my pile of papers. I'll keep you posted on that.

To answer johnqwertyful, no I cannot. I didn't learn it yet. That's what's making this problem tough :(
 
  • #11
Nope.
I'm stuck.
(ln3-(ln3)(5/3)^y)/(ln2-(ln3)(7/2)^y)
Same [0/0] story again.
 
  • #12
alingy1 said:
Nope.
I'm stuck.
(ln3-(ln3)(5/3)^y)/(ln2-(ln3)(7/2)^y)
Same [0/0] story again.

Prove a useful result first. As I asked before, what is lim x->0 (a^x-1)/x??
 
  • #13
1 evidently! :)
 
  • #14
After 4 hours of sweat, I got it. :) You guys/girls are awesome. :)
 
  • #15
That's not right. The limit is equal to 1 only if a=e.
 

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