Determine whether limit is indeterminate or has a fixed value

[songoku]

Homework Statement

Please see below

Relevant Equations

Indeterminate forms

Indeterminate forms are: $\frac{0}{0}, \frac{\infty}{\infty} , \infty - \infty, 0 . \infty , 1^{\infty}, 0^{0}, \infty^{0}$

My answer:
4, 9, 15, 17, 20 are inderterminate forms

1. always has a fixed finite value, which is zero

2. $0^{-\infty}=\frac{1}{0^{\infty}}=\frac{1}{0}=\infty$ so it never has a fixed finite value

3. always has a fixed finite value, which is one

5. $\infty . \infty=\infty$ so it never has a fixed finite value

6. $\infty^{-\infty}=\frac{1}{\infty^{\infty}}=\frac{1}{\infty}=0$ so it always has a fixed finite value

7. $\infty^{1}=\infty$ so it always has a fixed finite value

8. $\frac{\infty}{0}=\infty$ so it never has a fixed finite value

10. $\pi^{\infty}=\infty$ so it never has a fixed finite value

11. $1.\infty=\infty$ so it never has a fixed finite value

12. $1^{-\infty}=\frac{1}{1^{\infty}}$ , not sure about this one since $1^{\infty}$ is indeterminate form. My guess for this one is: $\frac{1}{1^{\infty}}=\frac{1}{1}=1$ so always has a fixed finite value

13. $\frac{1}{-\infty}=0$ so it always has a fixed finite value

14. $\infty^{\infty}=\infty$ so it never has a fixed finite value

16. $\infty^{-e}=\frac{1}{\infty^{e}}=\frac{1}{\infty}=0$ so it always has a fixed finite value

18. $\frac{0}{\infty}=0$ so it always has a fixed finite value

19. $\pi^{-\infty}=\frac{1}{\pi^{\infty}}=0$ so it always has a fixed finite value

Are my answers correct?

Thanks

 

[PeroK]

Your answer to 12 is not right.

 

[songoku]

Your answer to 12 is not right.

Would the answer be "indeterminate" because $1^{\infty}$ is indeterminate form?

Thanks

 

[PeroK]

Would the answer be "indeterminate" because $1^{\infty}$ is indeterminate form?

Thanks

Why not?

 

[songoku]

Why not?

I am not sure because in the explanation there is $\lim_{x \rightarrow \infty} 1^{x} = 1$ so I am confused whether it would be indeterminate form or 1

 

[PeroK]

I am not sure because in the explanation there is $\lim_{x \rightarrow \infty} 1^{x} = 1$ so I am confused whether it would be indeterminate form or 1

Your confusion would be resolved if you found an example where it is not $1$.

I fail to see how $1^{\infty}$ can be indeterminate, yet $\dfrac 1 {1^{\infty}}$ have a definite limit. That's a clear contradiction, is it not?

 

[pasmith]

I am not sure because in the explanation there is $\lim_{x \rightarrow \infty} 1^{x} = 1$ so I am confused whether it would be indeterminate form or 1

What about $\lim_{x \to \infty} (1 + kx^{-1})^x = e^{k}$? Is that of the form $1^{\infty}$?

 

[songoku]

Your confusion would be resolved if you found an example where it is not $1$.

You mean the one given in post #7?

What about $\lim_{x \to \infty} (1 + kx^{-1})^x = e^{k}$? Is that of the form $1^{\infty}$?

I think I get your hint

 

[WWGD]

Maybe a bit tricky, in that an indeterminate $\frac {\infty}{\infty}$ may be resolved by, e.g., L' Hopital's rule.

 

[songoku]

Thank you very much for the help and explanation PeroK, pasmith, WWGD