Leros
Lifer
The problem is:
lim (1/x^2)^x
x->0
y = lim (1/x^2)^x
ln y = ln lim (1/x^2)^x take ln of both sides
ln y = lim ln (1/x^2)^x move the ln inside the limit
ln y = lim[ x * ln(1/x^2) ] use exponential log property
I'm not sure where to go from here.
I said:
ln y = lim [ 0 *ln(1/x^2) ] = 0 since 0*x = 0; does l'hopital's rule apply?
ln y = 0
y = 1 this is the correct answer
Did I do this right? I am thinking not.
lim (1/x^2)^x
x->0
y = lim (1/x^2)^x
ln y = ln lim (1/x^2)^x take ln of both sides
ln y = lim ln (1/x^2)^x move the ln inside the limit
ln y = lim[ x * ln(1/x^2) ] use exponential log property
I'm not sure where to go from here.
I said:
ln y = lim [ 0 *ln(1/x^2) ] = 0 since 0*x = 0; does l'hopital's rule apply?
ln y = 0
y = 1 this is the correct answer
Did I do this right? I am thinking not.
