What is the Limit at Infinity for (2^x-5^x) / (3^x+5^x)?

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Homework Help Overview

The problem involves finding the limit as x approaches infinity for the expression (2^x - 5^x) / (3^x + 5^x), which falls under the subject area of limits in calculus.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants express uncertainty about how to approach the limit problem, with some suggesting dividing the numerator and denominator by a common term. Others discuss rewriting the terms in a different form to simplify the limit evaluation.

Discussion Status

Hints have been provided regarding rewriting the numerator and denominator, and some participants have begun to explore these suggestions. There is an ongoing exploration of different methods to approach the limit, but no consensus has been reached yet.

Contextual Notes

Participants are grappling with the challenge of handling exponential terms in the limit, and there is a mention of confusion regarding the simplification process. The original poster and others are seeking clarification on how to proceed with the problem.

terryds
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Homework Statement



30uqyoh.png

lim x->∞ (2^x-5^x) / (3^x+5^x)

Choices :

a. -1
b. -2/3
c. 1
d. 6
e. 25

2. The attempt at a solution

Hmmm.. I really have no idea about this.. This is an unusual problem..
Please tell me...
 
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terryds said:

Homework Statement



30uqyoh.png

lim x->∞ (2^x-5^x) / (3^x+5^x)

Choices :

a. -1
b. -2/3
c. 1
d. 6
e. 25

2. The attempt at a solution

Hmmm.. I really have no idea about this.. This is an unusual problem..
Please tell me...
Try dividing numerator and denominator by something.
 
Hint: nominator can be written as ##5^x((\frac{2}{5})^x-1)##. Also denominator can be written in a very similar way. Also i think you know that for any ##0<a<1## it is ##\lim\limits_{x \to +\infty}a^x=0##. If you use all this info i believe you should be able to find the correct answer.
 
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Raghav Gupta said:
Try dividing numerator and denominator by something.
Hmm..
I have no idea..
By what something ?
If I divide numerator and denominator by x, it will just make things more complicated
Since the 2^x/x can't be simplified more... (The bad thing is the x is the exponent, not in the number)
 
terryds said:
Hmm..
I have no idea..
By what something ?
If I divide numerator and denominator by x, it will just make things more complicated
Since the 2^x/x can't be simplified more... (The bad thing is the x is the exponent, not in the number)
Use Delta2 hints and try writing the denominator.
 
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Delta² said:
Hint: nominator can be written as ##5^x((\frac{2}{5})^x-1)##. Also denominator can be written in a very similar way. Also i think you know that for any ##0<a<1## it is ##\lim\limits_{x \to +\infty}a^x=0##. If you use all this info i believe you should be able to find the correct answer.

lim x->∞ 5^x((2/5)^x-1) / (5^x ((3/5)^x + 1))
lim x->∞ (2/5)^x - 1 / ((3/5)^x + 1)
-1/1 = -1

Okay, I've got that the answer is a. -1
Yeah, thanks for your hint :smile:
 
Notice that this was the same as "divide both numerator and denominator by 5^x".
 

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