# Help with limits problems

• Jan 18th 2010, 08:14 AM
Lord Darkin
Help with limits problems
My problem says, "there's a myth circulating among calculus students which states that all indeterminate forms of types $\displaystyle 0^0, infinity^0$ and 1^(infinity) have a value of 1 since anything to the zero power is 1 and 1 to any power is 1.

But the thing is, those 3 things I mentioned before are "descriptions of limits" rather than powers of numbers. Please show that such indeterminate forms can have any positive real value.

(PS: Apologize for the terrible typing, I don't know how to use latex well.)

1)

lim (as x-> 0+)
[x^(lna/(1+lnx))] = $\displaystyle 0^0$ = a

2)

lim (as x-> 0+)
[x^(lna/(1+lnx))] = infinity^0 = a

3)

lim (as x-> 0+)
[(x+1)^(lna/x)] = 1^(infinity) = a

I am really stumped on how to do these. I did #1 and ended up with 1/(a(x+1)) which means that it would end up with a fraction of 1/a if x->0, which does not equal a.

My attempt

x^(lna/(1+lnx))

I used the e^ln thing to get the exponent down. Then I used L'Hopital rule.

e^(1/(a(x+1)))

?
• Jan 18th 2010, 08:36 AM
Calculus26
See attachment

All are handled in basically the same way
• Jan 18th 2010, 09:23 AM
Lord Darkin
Thanks! It seems like my problem was that I changed ln(a) to 1/a.

Just wondering, now I know how to get y=a, how does the $\displaystyle 0^0$ thing relate to this problem?
• Jan 18th 2010, 09:26 AM
Calculus26
Quote:

lim (as x-> 0+)
[x^(lna/(1+lnx))] = http://www.mathhelpforum.com/math-he...b2aa2e4f-1.gif = a

as x -> 0 ln(x) -> -inf

ln(a)/-inf = 0

so you have 0^0 as x->0
• Jan 18th 2010, 10:31 AM
Lord Darkin
Looks good, I understand now since x = 0 would be the base and then the 0 explained in the last post would be the exponent.

How do I do this for part b and c where I need $\displaystyle \infty^0$ and $\displaystyle 1^\infty$?
• Jan 18th 2010, 11:14 AM
Calculus26
Quote:

2)

lim (as x-> 0+)
[x^(lna/(1+lnx))] = infinity^0 = a

in #2 the limit should be as x-> infinty not zero

It still evaluates to a

in #3 its obvious x+1 ->1
lna/x -> ln(a)/0 -> infinity
• Jan 18th 2010, 06:38 PM
Lord Darkin
That's weird, my problem definitely says that for all 3 parts the limit goes to 0. But I guess I'll explain that on the problem.

Anyway, looks like another problem solved. Thanks.
• Jan 18th 2010, 06:41 PM
Drexel28
Quote:

Originally Posted by Lord Darkin
My problem says, "there's a myth circulating among calculus students which states that all indeterminate forms of types $\displaystyle 0^0, infinity^0$ and 1^(infinity) have a value of 1 since anything to the zero power is 1 and 1 to any power is 1.

But the thing is, those 3 things I mentioned before are "descriptions of limits" rather than powers of numbers. Please show that such indeterminate forms can have any positive real value.

(PS: Apologize for the terrible typing, I don't know how to use latex well.)

1)

lim (as x-> 0+)
[x^(lna/(1+lnx))] = $\displaystyle 0^0$ = a

2)

lim (as x-> 0+)
[x^(lna/(1+lnx))] = infinity^0 = a

3)

lim (as x-> 0+)
[(x+1)^(lna/x)] = 1^(infinity) = a

I am really stumped on how to do these. I did #1 and ended up with 1/(a(x+1)) which means that it would end up with a fraction of 1/a if x->0, which does not equal a.

My attempt

x^(lna/(1+lnx))

I used the e^ln thing to get the exponent down. Then I used L'Hopital rule.

e^(1/(a(x+1)))

?

Are these set problems, or can you show that $\displaystyle 0^0$ can converges to any value?
• Jan 18th 2010, 06:42 PM
Calculus26
Quote:

That's weird, my problem definitely says that for all 3 parts the limit goes to 0. But I guess I'll explain that on the problem.

Anyway, looks like another problem solved. Thanks.

Note if that were the case 1 and 2 are identical
• Jan 23rd 2010, 02:34 PM
Lord Darkin
^Yes, they are exactly identical. ??? My teacher won't take questions for xtra credit (e.g. I can't ask why they are the same) so I'll just hand in what I have.

Thanks again to you guys.