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

AI Thread Summary
The limit at infinity for the expression (2^x-5^x) / (3^x+5^x) can be determined by rewriting both the numerator and denominator in terms of 5^x. By applying the hint to divide by 5^x, the expression simplifies to the limit of ((2/5)^x - 1) / ((3/5)^x + 1). As x approaches infinity, (2/5)^x and (3/5)^x approach zero, leading to the limit of -1/1. Therefore, the correct answer is a. -1.
terryds
Messages
392
Reaction score
13

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...
 
Physics news on Phys.org
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.
 
  • Like
Likes terryds
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.
 
  • Like
Likes terryds
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".
 
Back
Top