Question

describe the form of the limit. Does L'Hopital's rule apply?
a. 0/0
b. infinity/infinty
c. infinity - inifinity
d. infinity*0
e. 1 ^ infinity
f. zero^ zero
g. infinity ^ zero
h. the form limit is none of these

please remember to say yes or no if as to whether the rule applys ..thank you!

Answer 1

my educated guess. yes to only:
a, b, e

Answer 2

a and b

Answer 3

L'Hopital's Rule only applies to a form of 0/0 or infinity/infinity. Howeber, other forms can be changed to 0/0 or infinity/infinity. But the Rule can only be applied when it is of the form 0/0 or infinity/infinity. Some of your other forms are indeterminate forms, and can be changed to 0/0 or infinity/infinity. Some cannot, and may involve taking the ln of the expression, etc.

Answer 4

i think 1^inf as well

Answer 5

You cannot use direct L'Hopital's Rule with 1^infinity.

Answer 6

but you would have interest in turning it into a 0/0 or inf/inf form because it is identified as 1^inf like:
\[\lim_{x-> \infty} \left( e^{\frac{ 1 }{ x }} \right)^x\]

Answer 7

For 1^infinity, you have to take ln first.
For example, lim x->0 (1 + x)^(1/x)
Can't apply L'Hopital's Rule directly.
y = (1 + x)^(1/x)
Take ln on both sides and then apply L'Hopital's Rule.

Answer 8

Example; the function (sec x)^ (cot x) as x---->0 is of the form 1^infinity...but you cannot use a direct L'Hopital's Rule.

As I said earlier, you would try to convert it, if possible, to 0/0 or inifinity/infinity.

Answer 9

but you knew to that because it was 1^inf

Answer 10

So I agree with @ranga

Answer 11

my example is silly. e^(1/x)^x = e . lol

forget it

Answer 12

agree @Euler271

Answer 13

i agree

Answer 14

Classic example, \[\lim_{x \rightarrow \infty} (1+\frac{ 1 }{ x})^x\] appears to be 1^infinity, but is equal to e.