Question about indeterminate form

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Maged Saeed
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why $$1^\infty$$ is indeterminate form?
 
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Maged Saeed said:
why $$1^\infty$$ is indeterminate form?
1 raised to any finite power is 1, of course, but some limits are of this indeterminate form, and have a limit that isn't equal to 1.

The most famous example is this limit:
$$\lim_{x \to \infty}(1 + \frac{1}{x})^x$$

The base is approaching 1 and the exponent is "approaching" infinity. It can be shown that the value of this limit expression is the number e.
 
Thanks ,, I got it now
(:
 
PeroK said:
Anything with ##\infty## is an indeterminate form, because ##\infty## is not a number.

I don't believe that [itex]1/\infty[/itex] is an indeterminate form because its value cannot be anything other than 0.
 
Mentallic said:
I don't believe that [itex]1/\infty[/itex] is an indeterminate form because its value cannot be anything other than 0.
I agree. It's not indeterminate because you can determine what the limit will be.
 
PeroK said:
Anything with ##\infty## is an indeterminate form, because ##\infty## is not a number.
That's not true. While ##[\infty - \infty]## and ##[\frac{\infty}{\infty}]## are indeterminate forms, ##[\infty + \infty]## and ##[\infty * \infty]## are not considered indeterminate.
 
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Mark44 said:
That's not true. While ##[\infty - \infty]## and ##[\frac{\infty}{\infty}]## are indeterminate forms, ##[\infty + \infty]## and ##[\infty * \infty]## are not considered indeterminate.

Mark44 said:
I agree. It's not indeterminate because you can determine what the limit will be.

Me too .

How about
$$\frac{0}{\infty}$$

Is it indeterminate too?

I think so

:oldeyes:
 
Maged Saeed said:
Me too .

How about
$$\frac{0}{\infty}$$

Is it indeterminate too?

I think so

:oldeyes:
No, it is not indeterminate. If the numerator approaches 0 and the denominator becomes unbounded, the limit is 0.
 
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