# Solving Indeterminate Forms: Evaluating Limits with Exponential Functions

• alingy1
In summary: In general, it is equal to ln(a).In summary, the conversation was about evaluating the limit of (3^y-5^y)/(2^y-7^y) as y approaches 0. The participants were discussing different approaches and methods to solve the problem, including trying to cancel out y and using logarithm rules. Eventually, it was suggested to use the special limit lim x->0 (a^x-1)/x, which is equal to ln(a). After some confusion and mistakes, the final solution was reached and the participants expressed their gratitude to each other.
alingy1
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. :(

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?

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 :(

Nope.
I'm stuck.
(ln3-(ln3)(5/3)^y)/(ln2-(ln3)(7/2)^y)
Same [0/0] story again.

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??

1 evidently! :)

After 4 hours of sweat, I got it. :) You guys/girls are awesome. :)

That's not right. The limit is equal to 1 only if a=e.

## 1. What is the purpose of learning limits in basic Calculus I?

Limits are an essential concept in Calculus that help us understand the behavior of functions as they approach a certain point. They are crucial in understanding rates of change, continuity, and the derivative, which are fundamental concepts in Calculus.

## 2. How do you find the limit of a function?

To find the limit of a function, you can either plug in values that approach the desired point or use algebraic techniques such as factoring and simplifying. You can also use the graph of the function to visualize the behavior near the desired point.

## 3. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit is the limit of a function as it approaches a certain point from either the left or right side. It only takes into account the behavior of the function on one side of the desired point. A two-sided limit, on the other hand, considers the behavior of the function on both sides of the desired point.

## 4. How do you determine if a limit exists or not?

A limit exists if the left-sided limit and the right-sided limit are equal. In other words, if the function approaches the same value from both sides of the desired point, the limit exists. If the left-sided and right-sided limits are not equal, the limit does not exist.

## 5. Can you use limits to find the value of a function at a specific point?

No, limits do not give the actual value of a function at a particular point. They only tell us what the function approaches as it gets closer and closer to that point. To find the value of a function at a specific point, we need to evaluate the function at that point.

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