How would you go about answering this?

Question

If x and y are both positive evaluate
A.) lim= (ln(0.8x^p+2.0y^p))/p
p->0
B.) lim= (0.8x^p+2.0y^p)^(1/p)
p->0

Answer 1

For part A, we can use L'Hopital's rule:

lim [ln(0.8x^p + 2.0y^p)]/p
p->0

= lim [(0.8x^p + 2.0y^p)/(ln(0.8x^p + 2.0y^p))^-1]
p->0

Now, we can use L'Hopital's rule to evaluate the limit of the denominator:

= lim [(0.8x^p + 2.0y^p)/(-1)(0.8x^p + 2.0y^p)^(-2)(0.8ln(0.8x^p + 2.0y^p) + 2.0pyp^-1))]
p->0

= lim [-0.8ln(0.8x^p + 2.0y^p)/(0.8x^p + 2.0y^p)]
p->0

= -ln(0.8)/(x^0 + y^0)
(since lim (0.8x^p + 2.0y^p) = 0.8x^0 + 2.0y^0 = 2 > 0)

= -ln(0.8)/(x + y)

Therefore, the answer for part A is -ln(0.8)/(x + y).

For part B, we can rewrite the expression as:

lim [0.8x^p + 2.0y^p]^(1/p)
p->0

= lim e^(ln(0.8x^p + 2.0y^p)/p)
p->0

Now, we can use L'Hopital's rule again:

= lim e^[(ln(0.8x^p + 2.0y^p))^-1(0.8x^p ln(0.8x^p + 2.0y^p) + 2.0y^p ln(0.8x^p + 2.0y^p))]
p->0

= e^(0.8ln(0.8) + 2.0ln(2))
(since lim (0.8x^p + 2.0y^p) = 0.8x^0 + 2.0y^0 = 2 > 0)

= e^(ln(0.8^0.8 * 2^2))

= e^(ln(1.024))

= 1.024

Therefore, the answer for part B is 1.024.

Answer 2

I don't think the limit exists. Consider the case where y=0. Then
ln(0.8x^p)/p = (ln0.8 + p*lnx)/p = ln0.8/p + lnx
As p→0, lnx→-∞ and 0.8/p→∞
And if y is not zero, then you have yet another number divided by p, which will also be undefined.

Answer 3

You are correct. I apologize for my mistake in the previous answer. As you have pointed out, if y=0, then the limit does not exist. Additionally, if y is not zero, then we cannot apply L'Hopital's rule to evaluate the limit. Therefore, the limit is undefined in both parts A and B. Thank you for bringing this to my attention.