Solving Limits - Questions on x^(sinx) & (9^x)/(8^x)

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The limit of x^(sinx) as x approaches 0 from the positive side is confirmed to be 1. For the second limit, (9^x)/(8^x) simplifies to (9/8)^x, which is greater than 1. As x approaches infinity, (9/8)^x diverges to infinity since the base is greater than 1. Understanding the behavior of exponential functions based on their bases is crucial for solving such limits. The discussion emphasizes the simplicity of these limit calculations when the properties of exponents are applied correctly.
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Hello, I have two questions concerning limits

1) lim x-> 0+ x^(sinx)
2) lim x-> +inf. (9^x)/(8^x)

The first one gives me 1 (e^0 = 1) .. is that correct?
The 2nd one I don't know how to do. Can someone please explain the 2nd one for me?
Thanks a lot
 
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1) Correct

2) (9^x)/(8^x) = (9/8)^x and 9/8 > 1. How does a^x behave if a<1, a=1, a>1 ?
 
Gokul43201 said:
1) Correct

2) (9^x)/(8^x) = (9/8)^x and 9/8 > 1. How does a^x behave if a<1, a=1, a>1 ?


I knew it was that simple! Thanks for the help :)
 

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